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September 15

having problems with conversion of measurements

somethings i can do but this is not one of them. here are some examples.15 meters to millimters?3.5 tons to pounds?6800 seconds to hours?could someone please help me understand how you change these? —Preceding unsigned comment added by 71.185.146.144 (talk) 09:14, 15 September 2008 (UTC)

I put them in google and it does them for me...

http://www.google.co.uk/search?hl=en&q=15+meters+in+millimeters&meta= http://www.google.co.uk/search?hl=en&q=3.5+tonnes+in+pounds&meta= http://www.google.co.uk/search?hl=en&q=6800+seconds+in+hours&meta= Simple as that. Of course without google things get a bit harder... 194.221.133.226 (talk) 10:35, 15 September 2008 (UTC)

To change between units you need to know how many of the one unit are in the other unit. Then you multiply by how many of the first unit you have. Say that there are 5 "apples" in a "bag" and you have 2.4 bags, then you have 2.4 * 5 = 12 apples. To go the other way, there is 1/5 = 0.2 bags per apple, so 9 apples is 9 * 0.2 = 1.8 bags. So all you need to know is how many millimetres in one metre, how many pounds in one ton, seconds in an hour etc. If you don't know these from memory then the relevant wikipedia articles will help you. -- SGBailey (talk) 11:28, 15 September 2008 (UTC)

Annual Percentage Rate

So I am in a confusion about what “APR” is. In particular, in the simplest case (ignoring everything but the principal) whether it is a nominal or effective rate.

The article annual percentage rate claims it is an effective rate. Through some online searching I’ve found conflicting sites that seem to say opposite things. Eventually I dug out my old theory of interest book (Theory of interest by Kellison) which said it was a nominal rate. Asking a friend, she pulled out hers (Mathematics of interest rate and finance by Guthrie and Lemon), which also says it is a nominal rate.

So I just wanted to check with anyone that might know more about this than me to see if the article is actually wrong, before changing anything (or telling the students in 101 my class something incorrect next week!) GromXXVII (talk) 12:53, 15 September 2008 (UTC)

I'm not going by the article, I have a Ba in finance, and graduated with high honors from the best college in the nation. The APR is not an effective rate. Here's two examples to clear things up.

A $100 bond pays a $3 coupon every 6 months.

The APR is 6.000%

The effective interest rate on the bond is 6.090%

A $100 bond pays a $4 coupon, every 12 months

The APR is 4.000%

The effective interest rate on the bond is 4.000%

I'll answer any further questions, and will provide relevant sources, at your request. The following example may be confusing, but here it goes anyway.

A credit card has a 29.9% APR

If I charged $200, and pay off $100 of it, then I'll have a $100 balance remaining. When my new bill comes in, I'll owe $102.492

The effective interest rate is 34.358%, and if you figure out how I computed this, you'll have came away with full understanding, as every assumption is reflected. Sentriclecub (talk) 13:11, 15 September 2008 (UTC)

Surely a $100 bond that pays $4 each 12 months is not a 6% rate? The credit card example is also flawed, in that many cards compute interest based on an average daily balance, thus not knowing when the partial payment was made makes it impossible to computer the next month's finance charge. Offering sources "upon request" is redundant, the request is implicit in the use of the reference desk. --LarryMac | Talk 13:24, 15 September 2008 (UTC)

Thanks fixed it, ask me how to derive a physics formula, no problem. Trying to multiply 4x1, that's a challenge. The credit card example isn't flawed, if you reread the stipulation that I started the month with a $100 remainder balance and made no new charges. The reason I don't want to offer sources, is to avoid the conundrum that its a real possibility that some sources may calculuate an EAR and APR in different ways. If I gave a source, I would be forced to say that the wikipedia article is wrong, which I don't believe. I would have given the source right away, but I want to compare my author's accuracy to the wikipedia article's accuracy. Secondly, I will point out that you are correct about when the payment is applied, however the example implies in my favor. Since I'm providing clearification, I don't want to give more information than necessary, but I also agree on your point that there is risk resulting with not specificating every detail such as was it februrary on or off leap year, or was it a month with 31 days, etc... Upon further inspection, the credit card example is not flawed. The rate per period is APR/n. LarryMac, may I respectfully ask if you have studied this stuff in school? Something is intriguing me, a subtle mistake you've made in your post. I take it you are a math expert in another area? I don't want to come across bad, I'm representing my school now, and I hope my behavior today is helpful and respectful. I'm 24, and one of the worst feelings I ever had on wikipedia was upsetting someone at the ref desk. So please be patient with me, and my only intention here today is to help answer this guy's question, while Primum non nocere to the other helpers. Sentriclecub (talk) 14:02, 15 September 2008 (UTC)

I understand the effective interest rate in your credit card example being 34.358%=(1+0.299/12)^12. But I’m still a little confused in that APR then appears to be a nominal interest rate (in that I could divide by 12 to get the monthly interest charged of 2.49%), but you said the Wikipedia article isn’t wrong? The first line is “Annual percentage rate (APR) is an expression of the effective interest rate” which seems to say the opposite. (as far as I know it’s not possible to be both an effective and nominal rate unless if the frequency of compounding is the period). GromXXVII (talk) 23:02, 15 September 2008 (UTC)

Yeah, I discussed that earlier over at a discussion page somewhere. Here is the quote--The first sentence is already prone to having people miseducated. Annual percentage rate (APR) is an expression of the effective interest rate the borrower will pay on a loan which is accurate, but completely misleading, because the APR is not an effective interest rate. It is an expression for one, but not one itself. -/quote I spoke to some of the editors in finance articles, and they're okay with me saying that the line is wrong. I just didn't want to offend anyone over there. APR is a nominal rate, (and is an effective rate if and only if n=1). I am going to work with Wikipedia:FINANCE to help fix inaccurate statements in existing prominent articles. The finance articles on wikipedia are in poor shape (maybe 5 or so major ones are rife with errors) so I'm going to volunteer some time over there, even though I'm going to med-school and leaving finance behind me. It's where I'm needed the most, so I'll try and make myself compare articles to my textbooks and rule in favor of my book. At least that way, APR and the articles which link to APR will all be consistent. If someone else wants to use a source that APR is an effective interest rate, they can come behind me--I dont care. But as it stands now nominal interest rate, effective interest rate, real interest rate and annual percentage rate contradict, so I already have the permission from the wikiproject over there, to do a mass clean-thru of the pages. My 5 finance professors and all of their textbooks (5+5=10) are all consistent with each other, and I'll source all my texts, so in a few days, the articles will be clear and those negative tags will come off the tops of all 6 pages that I plan to work on. Sentriclecub (talk) 00:04, 16 September 2008 (UTC)

Ahhh. I see. I was erroneously taking “is an expression of” to mean “is”. You have my thanks in advance for any clarifications you’re able to make on those pages. GromXXVII (talk) 00:17, 16 September 2008 (UTC)

addendum 1

APR = (Per-period rate) X (Periods per year)

EAR = $e^{APR}-1\,$

APR = $ln(1+EPR)\,$

I'm going to look more into all this for you, and will have a full report back in an hour--it looks as though neither the APR nor EAR are independent of the number of compounding periods. The best way to interpret interest rates is with calculus. It appears that the APR and EAR are dependent on the number of periods per year, which bothers me. I'll re-read everything, and will also consult my HP-10Bii user manual, 2 feet away from me at all times. While you are waiting, check out Khan academy on youtube, this guy is a bona fide expert. The finance videos he has produced are wonderful. Sentriclecub (talk) 13:35, 15 September 2008 (UTC)

To clarify, the EAR is the Effective Annual Rate and in the formula above, the APR is the force of interest or the continuously compounded rate of interest. Zain Ebrahim (talk) 13:43, 15 September 2008 (UTC)

It's both nominal and effective. The opposite of nominal is real. See nominal interest rate, real interest rate and effective interest rate. --Tango (talk) 12:59, 15 September 2008 (UTC)

From reading the above definitions I would say it is always effective but only nominal if the interest period is annual (once a year). -- Q Chris (talk) 13:09, 15 September 2008 (UTC)

For this purpose, you need to look at the second definition at nominal interest rate, in which case a rate can't be nominal and effective. When I studied financial maths, the lecturers intentionally avoided the term APR because it changes from person to person. Zain Ebrahim (talk) 13:12, 15 September 2008 (UTC)

Actually, it depends where you are in the world - see annual percentage rate#Region-specific details. In the US, APR is defined as periodic interest rate times the number of compounding periods in a year, so it is a nominal interest rate. In the UK, however, APR is the effective interest rate, which factors in compounding and is not the same as the nominal rate unless the interest is only compounded once a year (in the US the effective rate is called the Annual Percentage Yield or APY - see this Investopedia article). So a loan with monthly interest payments of 0.5% per month would be described in the US as having an APR of 6%, but in the UK its APR would be 6.17% (this example is used in the nominal interest rate article). As in many instances of financial terminology, US and UK just follow different rules. Gandalf61 (talk) 13:27, 15 September 2008 (UTC)

Here are some sentences straight out of my book. Returns on assets with regular cash flows, such as mortgages and bonds, usually are quoated as an APR. The APR can be translated to an EAR by remembering that APR = (Per-period rate)x(Periods per year). Therefore to obtain the EAR if there are n compounding periods in the year, we solve the equation $1+EAR=(1+{\frac {APR}{n}})^{n}$. Sentriclecub (talk) 14:09, 15 September 2008 (UTC)

So for n large, 1+EAR is approximately e^APR. Algebraist 14:23, 15 September 2008 (UTC)

Here are some sentences straight out of my book. Returns on assets with regular cash flows, such as mortgages and bonds, usually are quoated as an APR. The APR can be translated to an EAR by remembering that APR = (Per-period rate)x(Periods per year). Therefore to obtain the EAR if there are n compounding periods in the year, we solve the equation $1+EAR=(1+{\frac {APR}{n}})^{n}$.

Additionally, I'm glad to see the point made that different sources say different things. I personally think any interest rate which is dependent on the number of periods per year, is a flawed measure. So I think with this information, we can call it a closed case. Finance has a way of trying to simplify formulas which are inherently exponential into algebraic. This is why the APR is probably best described as a nominal rate, not as an effective rate. Since the APR is usually divided into 12 periods per year (afterall, people make mortgage payments, and credit card payments once per month, not once per 30 days) so its good for consumers ultimately. Afterall, what if a homeowner took out a 8% home equity loan to pay a 9% credit card, and one of the APR's was calculuated assuming 12 periods per year, and the other APR was confounded and ended up losing EV on the deal. So with this reasoning, (and in my education, I studied the interest rate schemes which lead to federal standards--probably in the APR article) therefore I would argue that the APR is a nominal interest rate, since the reason the government makes Rent-a-Center publish an APR is so that consumers know they can compare it apples to apples like a mortgage payment, which they are familiar with.

Okay, even further into the philosophy of all this, I have finally concluded that the APR is a number to help consumers from getting duped. Say someone goes to rent-a-center and buys $800 rims, for their car and finances it. That guy can ask virtually anyone if 18% APR is a good deal, or not. He's not actually interested in his monthly payment, or in computing the principle to interest ratio of each minimum monthly payment. He just wants to know if the interest rate, if he's getting ripped off on the financing part of the purchase. In other words, I conjecture that an APR is a way for consumers to know if they are being taken advantage of, under the financing terms. As an anecdote, I heard with my own ears, a Ron Legrand audio lecture, on how to purchase a stream of cash flows from a landlord. Suppose the mark goes up to the landlord (who is a landlord, uses seller-financed a mortgage to the buyer) and if the principle on the mortgage is worth $160,000, make the landowner an offer to buy all the cash flows from now until when the principle on the mortgage is $120,000. You can play around with some of these numbers, but the mark ends up "buying" the $40,000 difference at a negotiated price. These are similar to the Carlton Sheets real estate courses. You'd be amazed at how much profit this comes out to be for the mark. So, I believe that APR is a nominal measure, which most consumers in america know that the default rate on a credit card is extremely undesirable. How do they know that a 30% APR is undesirable? Because APR is a nominal number which is widespread (all mortgages and credit cards I believe are federally required to report APR) so that consumers don't have to take a finance course to understand that banks borrow money at a very low rate, around 4%, and that people with good credit can borrow at around 8%, and that you should only finance at 18% APR, rims which cost $800 if you wish to take on a undesirable interest rate. Therefore, you should only buy them if you make enough utility on the $800 price, to justify the exorbitant interest rate. This would be the case if the rims were of some special case of tremendous value, like maybe a clearance sale. My closing point, which I hope I've made, is that you should ask your self where APR's came from? What is the history of APR's. They are designed to help consumers understand (through the easy to understand simple interest method) a complex formula. Most consumers aren't good at calculating exponents and natural logarithms, but almost anyone that can get a loan, understands how to divide by 12 and multiply by 12. Thus APR's are understood by the majority of consumers. Consumers don't understand EAR in the same proportion as those that would understand simple interest. Therefore the APR which is most helpful, is the one most everyone can understand. APR is nominal. To calculuate the interest for one month, you divide the APR by 12. This is within the abilities of the largest proportion of consumers. Then again, I also question if credit card lobbyists may have been behind this whole enigma. Afterall, it lets them quote an APR of 29.900, but if this value were computed using exponents and non-simple-interest, they would have to report 34.358. But I think its that the greatest proportion consumers understand simple interest. The government (that regulates how consumers are told the interest rate) will guide corporations to quote an interest rate which people will understand. Sentriclecub (talk) 14:42, 15 September 2008 (UTC) Also, I went back and read the APR article and the third sentence is APR is intended to make it easier to compare lenders and loan options. which summarizes my point.

combinatorics question

A bridge club has ten members. Four get together to play everyday. Show that in two years one group of four people gets together at least four times.

The only thing that I can think of is that in the first 211 days one group definitely gets together twice. Any other ideas?--Shahab (talk) 19:03, 15 September 2008 (UTC)

Your argument uses the basic version of the pigeonhole principle, but there's also a generalized version, pointing out that since 3×210 = 630 days is the longest possible period of time without any group playing more than three times, in 631 days some group must play four times. -- Jao (talk) 19:36, 15 September 2008 (UTC)

Thank you. Here's another problem which is troubling me. I can't figure out how to apply the pigeonhole principle here: Prove if the numbers 1, 2, 3, . . . 12 are randomly placed around a circle, there must be three consecutive numbers whose sum is at least 19.--Shahab (talk) 07:00, 16 September 2008 (UTC)

Going round the circle, there are 12 sums of three consecutive numbers. You can work out the total of these 12 sums (each number on the circle is counted three times, in three different sums), so you can work out the average value of these 12 sums. The 12 sums cannot all be less than their average, so at least one sum must be greater than or equal to their average. Gandalf61 (talk) 09:23, 16 September 2008 (UTC)

As far as I understand this the total of the 12 sums must be 234, which when divided by 12 would give 19.5 proving more then asked for. Clearly this is an elegant solution. Thanks. But I want to explicitly involve the the pigeonhole principle in the solution coz the problem occurs in that section in my book. (Another way that I thought of was that if each of the sums is less then or equal to 18 then their total can be at most 12 times 18 i.e. 216, contradicting the fact that the total is 234).--Shahab (talk) 09:46, 16 September 2008 (UTC)

Divide the circle into groups of three. If each number occurs once there will be 4 such groups. By the generalized pigeonhole principle if none of these groups sum to more than 18 then the whole circle can not sum to more than 18*4=72, but 1+2+...+12=78.Taemyr (talk) 14:21, 16 September 2008 (UTC)

Another approach, less combinatoric but more direct: try to create an arrangement in which each sum of 3 consecutive numbers is 18 or less. Note that no two out of 9, 10, 11 and 12 can be in the same sum, so no two of these four can be neighbours or have just one number between them. So your arrangement must be a cyclic permutation of abbabbabbabb where the as are 9, 10, 11 and 12. Now show that there is nowhere left to put 8 without creating a sum of 19 or greater. Gandalf61 (talk) 09:33, 16 September 2008 (UTC)

Population and area question

Population and area question

How many times greater is the population of China than Australia? Secondly, how many times greater is the land area of China than Australia? —Preceding unsigned comment added by 81.151.147.129 (talk) 19:17, 15 September 2008 (UTC)

You can certainly find the relevant figures in our articles on Australia and People's Republic of China. -- Coneslayer (talk) 19:29, 15 September 2008 (UTC)

Word Problem- how to setting up the equations

Two pipes feed into a tank. The large pipe can fill the tank in 2 hours, the smaller pipe can fill the tank in 6 hours. If both pipes are used together, how long would it take to fill the tank.

Thank you for your help. —Preceding unsigned comment added by Evansranch (talk • contribs) 20:20, 15 September 2008 (UTC)

A good way to start is by answering the question: "It would take x hours." Then think about it: how many tanks will the large pipe fill in x hours? How many tanks will the small pipe fill? How many will they fill together? -- Jao (talk) 20:29, 15 September 2008 (UTC)

Per hour, the large pipe fills half the tank, the small one one sixth of it. So what fraction together? How many hours to make this fraction into one, i.e. a full tank?—86.148.186.156 (talk) 21:42, 16 September 2008 (UTC)

In one hour, the 'faster' pipe can fill 1/2 of the tank and the 'slower' pipe can fill 1/6 of the tank. (1/2 + 1/6) = 4/6 = 2/3 of the tank. Suppose it takes 'x' hours to fill the tank. Then 2x/3 = 1 which implies that x = 3/2 hours. Therefore the tank is filled in 1 hour and 30 minutes.

Topology Expert (talk) 13:15, 17 September 2008 (UTC)

We usually avoid actually give the answer to questions like this. It's much better to guide the OP towards getting the answer themselves than just giving it. --Tango (talk) 17:29, 17 September 2008 (UTC)

Combinations

A question I am working on reads:

A bag contains 20 chocolates, 15 toffees and 12 peppermints. If three sweets are chosen at random what is the probability that they are:

a) All different,

b) All chocolates,

c) All the same,

d) All not chocolates?

I have calculated the answer to a, to be 0.222 (which is correct), using the following steps:

47C3 = 16215

Chocolates * Toffees * Peppermints = 3600

3600/16215 = 0.222

Firstly, while I know this works, I do not fully understand why, and I am completely stuck on all the other sections. Can anyone point me in the right direction? Thanks. --AFairfax (talk) 20:39, 15 September 2008 (UTC)

What you've done is calculate the number of possible choices of three sweets, that's the 47C3, and the number of choices satisfying the condition that they all be different, the 3600. The reason you can just multiply the number of each type together is because you know there must be one chocolate, one toffee and one peppermint, so the only choice is in which of the chocolates, which of the toffees and which of the peppermints your choose. You then divide one number by the other to get the probability. Does that make any sense? The other questions are answered the same way - you need to calculate how many different ways you can satisfy each condition. So for (b), you need to work out how many ways you can choose 3 chocolates from a bag containing 20 chocolates (you can ignore the other sweets since they aren't allowed) - that's just 20C3. I'll let you try (c) and (d). Good luck! --Tango (talk) 21:11, 15 September 2008 (UTC)

The annoying thing is I already knew all that really, just didn't put it into practice in the correct way. Thanks for the pointer. --AFairfax (talk) 22:17, 15 September 2008 (UTC)

There are definitely 16215 ways of drawing three sweets and 3600 ways of drawing three sweets of different kinds. But when you divide the one by the other to get a probability you're introducing a new assumption: that all 16215 draws are equally likely. It's a standard assumption in word problems like this, and it's probably meant to be implied by the phrase "chosen at random", but it would be tricky to achieve in reality with a heterogeneous collection like this one. Shaking the bag and then letting one candy fall out would probably favor some types over others by a significant margin. So always think carefully before you treat a ratio of integers as a probability. -- BenRG (talk) 23:55, 15 September 2008 (UTC)

(edit conflict)

$20\cdot 15\cdot 12/{\tbinom {47}{3}}=20\cdot 15\cdot 12\cdot {\frac {3!44!}{47!}}=20\cdot 15\cdot 12\cdot {\frac {3!}{45\cdot 46\cdot 47}}={\frac {20}{47}}\cdot {\frac {15}{46}}\cdot {\frac {12}{45}}\cdot 3!$

Explanation of the derived formula: First we choose among 20 chocolates out of 47 sweets, second we choose among 15 toffees out of 46 remaining sweets, third we choose among 12 peppermints out of 45 remaining sweets. 3! is the number of possible orders of selections (chocolates, toffees, peppermints; or chocolates, peppermints, toffees; or toffees, peppermints, chocolates etc.). According to mathematics of outs in Texas Hold'em, b) is determined by formula ${\frac {20}{47}}\cdot {\frac {19}{46}}\cdot {\frac {18}{45}}=0.070$.

Similarly, c) ${\frac {20}{47}}\cdot {\frac {19}{46}}\cdot {\frac {18}{45}}+{\frac {15}{47}}\cdot {\frac {14}{46}}\cdot {\frac {13}{45}}+{\frac {12}{47}}\cdot {\frac {11}{46}}\cdot {\frac {10}{45}}=0.112$

And d) ${\frac {47-20}{47}}\cdot {\frac {46-20}{46}}\cdot {\frac {45-20}{45}}=0.180$

--Admiral Norton ^(talk) 21:19, 15 September 2008 (UTC)

This is a Hypergeometric distribution.

Topology Expert (talk) 13:10, 17 September 2008 (UTC)

ms^-3

how would you describe the concept of m/s/s/s? 84.13.30.238 (talk) 20:41, 15 September 2008 (UTC)

• Jerk. --PaulTaylor (talk) 21:00, 15 September 2008 (UTC)

  Nitpick: The units of jerk. Zain Ebrahim (talk) 22:02, 15 September 2008 (UTC)

Thanx Paul and Zain! I was just curious... thnx again ATMarsden^{Talk · {Semi-Retired}} 15:50, 16 September 2008 (UTC)

m/s/s means the change in m/s per second; or the change in velocity per second or acceleration. Therefore, m/s/s/s means the change in m/s/s per second; or the change in acceleration per second. This has no particular name but the third derivative of a function represents this. —Preceding unsigned comment added by Topology Expert (talk • contribs) 13:09, 17 September 2008 (UTC)

As already noted, it has a name: jerk. -- Jao (talk) 19:08, 17 September 2008 (UTC)

Gauss and triangular numbers

from Carl Friedrich Gauss:

"...every positive integer is representable as a sum of at most three triangular numbers..."

A list of triangular numbers taken from triangular number:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

I know I'm missing something here. Integers are in the set {..., -2, -1, 0, 1, 2, ...}, and 1, 3, 6, and 10 are all triangular numbers. 1 + 3 + 6 + 10 = 20 and 20 is an integer. Therefore, the sum of four triangular numbers is equal to an element in the set of integers. What am I missing? Ζρς ι'β' ^¡hábleme! 22:27, 15 September 2008 (UTC)

20 = 10 + 10? Confusing Manifestation(Say hi!) 22:48, 15 September 2008 (UTC)

And just to clarify, what Gauss found was that given any positive integer, you can break it down into a sum of three or fewer triangular numbers. Not that adding more than three triangular numbers together suddenly throws you out of the set of integers entirely. Confusing Manifestation(Say hi!) 22:49, 15 September 2008 (UTC)

Ok, thanks. That's what I was missing. 67.54.224.199 (talk) 23:15, 15 September 2008 (UTC)

Or since 0 is a triangular number then 20=10+10+0 with no change to the statement. Dmcq (talk) 08:19, 16 September 2008 (UTC)

The statement says "at most three", so there is no need to pad it out with zeros. --Tango (talk) 10:44, 16 September 2008 (UTC)

September 16

Constructible numbers

The Wikipedia article on constructible numbers states that :

If z is constructible, then it is algebraic, and its minimal irreducible polynomial has degree and power of 2, or equivalently, the field extension Q(z)/Q has dimension and power of 2. One should note that it is true, (but not obvious to show) that the converse is false — this is not a sufficient condition for constructibility. However, this defect can be remedied by considering the normal closure of Q(z)/Q.

Can anyone tell me how to construct such a number ? Even without an explicit construction, a Galois theoretic explanation of what type of polynomial this number should be a root of.

My understanding is that you need to find a number α such that [Q(α)/Q]=2^k but that, if E is the normal closure of Q(α), then [E/Q]!=2^k for any k. How is that possible ?

I thought that, as [Q(α)/Q]=2^k, there must be an irreducible polynomial P(x) in Q[x] of degree 2^k, of which α is a root. P(x) can't split over Q[x]/(P(x)), otherwise the extension would be normal. But all roots of P(x) must be of degree 2^k so I don't see how we could get a normal closure with degree over Q which isn't a power of two.

So, where does such a non-constructible number α with [Q(α)/Q]=2^k come from ?

Thanks. --Xedi^Talk 05:05, 16 September 2008 (UTC)

I believe the article is just saying that having the highest power of your polynomial a power of two isn't sufficient, for instance x in x⁴+x+1=0. However if you extend the rationals with the roots of this equation you'll get cube roots as well coming up. Dmcq (talk) 08:43, 16 September 2008 (UTC)

By the way if you like this then you might like Mathematics of paper folding where trisection of angles can be done. Dmcq (talk) 08:46, 16 September 2008 (UTC)

Well, the thing is, x⁴+x+1 is an irreducible polynomial of degree 4, so all its roots must be of degree 4 over Q, no ? And then, taking normal closure is not going to affect this degree, so I don't see how "the defect can be remedied" that way. I did have the impression something could go wrong if cube roots were involved somewhere, but I don't see how taking normal closure remedies this. --Xedi^Talk 16:10, 16 September 2008 (UTC)

In general, if p is an irreducible degree n polynomial over Q, then any root of p generates a n-dimensional extension. The normal closure of any of these extensions is the full splitting field of p. The degree of this splitting field is divisible by n and divides n!. In this case, the degree of the splitting field is 24. Algebraist 16:27, 16 September 2008 (UTC)

So, at which point does the factor of 3 in 24 come in then ? And how do you know that in this case the degree of the splitting field is 24 (Other than just giving the order of the Galois group of the extension) ? --Xedi^Talk 16:50, 16 September 2008 (UTC)

I calculated it using standard tricks for computing Galois groups using reduction mod m. The factor of 3 comes in because when we adjoin a single root of the polynomial, this splits the poly into a product of a linear poly and an irreducible cubic, and we have to adjoin a root of that cubic to make the poly split further. For a simpler example, consider x³-2. This is irreducible, so we adjoin a root a. The polynomial then splits as (x-a)(x²+ax+a²). The second factor is irreducible; to split it, we need to adjoin a square root of -3. Thus the splitting field has degree 6. Algebraist 17:15, 16 September 2008 (UTC)

Ah yes ok. Thanks a lot ! --Xedi^Talk 17:21, 16 September 2008 (UTC)

Baffling Proportions Problem

The other day I was helping my younger brother with a maths problem. His teacher had given him the problem in a quiz. "Working 8 hours a day, 24 men do a certain task in 15 days. Find how many days it would take 16 men working 9 hours a day to do the same task"

It is a inverse proportion problem but there are 5 pieces of information here. Usually these kind of problems involve 3 pieces of information. This problem has got me very confused and I am feeling pretty embarrassed. Would be glad if I could get some help here! Thank you! —Preceding unsigned comment added by 203.81.220.213 (talk) 16:33, 16 September 2008 (UTC)

Try breaking it all down into man-hours (the amount of work done by one man in one hour). Use the first scenario to work out how many man-hours the task requires, and then use that to work out the time required for the second scenario. --Tango (talk) 17:23, 16 September 2008 (UTC)

The 5 pieces of information fall into one base piece (the given time to do the task, i.e. 15 days), and two pairs of adjustment pieces (the number of men, and how many hours they work each day), where each pair forms a fraction to multiply the base piece. So the fraction for men is either 24/16 or 16/24, and for hours either 8/9 or 9/8. Which fraction to use in each case will come by considering the direction of change to the base figure resulting from (a) fewer men (b) more hours per day. When that has been decided, the answer will come by multiplying the original time by the two fractions in turn. This approach lends itself to there being more than two factors in the problem, say a rate of efficiency as well. If the 24 men worked at 80% and the 16 men at 90%, there would just be a third fraction (80/90 or 90/80, to be decided) in the multiplicative chain.—86.148.186.156 (talk) 22:15, 16 September 2008 (UTC)

Random trivia: hapless schoolboys once solved these problems using the "double rule of three," immortalized in Lewis Carroll's Mad Gardener's Song:

He thought he saw a Garden-Door that opened with a key:

He looked again, and found it was a Double Rule of Three:

‘And all its mystery,’ he said, ‘Is clear as day to me!’

The "mystery" was a reference to the fact that this rule was famously difficult to teach. I don't recommend teaching it to your brother—far better to teach general problem-solving techniques and real understanding of the problem. But here's an explanation of the rule from an 1830 textbook. It's interesting to note that the word problem they chose as their example can't correctly be solved this way—if $100 gains $6 in one year, it's not likely to gain $6n in n years. I wonder if any of the students ever noticed that. For some real soul-killing awfulness, check out this page. "Three sailors having been aboard 9 months 1/4, received 40ℓ 3/15; I demand how much 100 sailors must receive for 28 months 3/7 service? Answer 4118ℓ 6s 0d 1qrs 1305/11655." -- BenRG (talk) 23:13, 16 September 2008 (UTC)

Is there really any 'problem' in this?

Input data: "Working 8 hours a day, 24 men do a certain task in 15 days."

Twenty-four men do the work, so each of them does a 24-th part of the work in 15 days. He works 8 hours per day, so he completes his 1/24 of work in 15×8=120 hours — that means he does a 1/120 part of his work per hour, so one worker does 1/120×1/24 = 1/2880 of the task per hour.

Question: "Find how many days it would take 16 men working 9 hours a day to do the same task"

Each man, working 9 hr/day, does 9×1/2880 = 1/320 of the task per day. Sixteen workes do 16×1/320 = 1/20 of the taks a day. So they need 20 days to complete the work. Is anything obscure here? --CiaPan (talk) 07:19, 17 September 2008 (UTC)

This is a problem which requires more logical thinking than calculations. The answer from my calculations is: 23 days, 7 hours and 30 minutes. Try to see whether you get the same answer. Topology Expert (talk) 12:56, 17 September 2008 (UTC)

Nope. But this way is non-baffling (as outlined by Tango above):

• 8 hours/day $\times$ 15 days $\times$ 24 men = 2880 man-hours of work to be done (calculated by doing it the original way).
  
• 9 hours/day $\times$ 16 men = 144 man-hours/day available (by doing it the new way)
  
• 2880 man-hours (of work to be done) $\div$ 144 man-hours/day (by doing it the new way) = 20 days. -hydnjo talk 13:00, 17 September 2008 (UTC)

Huh? 'but this is non-baffling'...??? I apologize then, must have missed something important. I understand 'problems' are sometimes given in a complex, sophisticated form, but thought the solution should always be made as simple as possible, not baffling. At least that's what I remember from schools. --CiaPan (talk) 14:03, 17 September 2008 (UTC)

Yes, and Hydnjo's solution (which is the one I recommended at the top of this section, although I didn't actually give the answer) is simpler than yours. At least, it is to me. --Tango (talk) 17:27, 17 September 2008 (UTC)

Yeah, I spelt it out because no one seemed to be picking up on your simple explanation. -hydnjo talk 23:17, 17 September 2008 (UTC)

September 17

Functions

I'm given a graph, however I am confused about how to solve something like $2f(x+3)=52$ algebraically for x. Could someone please point me in the right direction. (By the way, this is a homework question.) I have to show my work, or I would just use the graph. TIA, 67.54.224.199 (talk) 01:45, 17 September 2008 (UTC)

If you are given an actual formula for f, then you can solve algebraically, starting with appropriate substitution. For example, if you were given f(x) = x^2, then f(x+3) = (x+3)^2 and from that you then solve 2 (x+3)^2 = 52. If you are only given a graph, then you would need to first use the graph to find an approximate solution to 2 f(x') = 52, and then using that approximate solution to solve x' = x + 3 for x. Confusing Manifestation(Say hi!) 03:50, 17 September 2008 (UTC)

In more details, locate the intersections of your graph and the horizontal line y=26 (52/2) to find x' = x+3. Then x = x'-3 are your solutions if there are more than one point in the intersection. twma 10:08, 17 September 2008 (UTC)

Need x so that: f(x+3) = 26. Therefore, you find x' such that f(x') = 26 and subtract 3 from x' to get the desired x (since f(x+3) = f(x'-3+3) = f(x') = 26 as desired).

The most important thing to remember is that f(x-a) translates the graph a units to the right (to get the same value of f(x), you need to add 'a' to the x-value). Similarly, f(x+a) translates the graph 'a' units to the left (to get the same value of f(x), you need to subtract a from the x-value so the graph must be translated 'a' units left). Similarly, -f(x) reflects f about x-axis and f(-x) reflects f about y-axis. In general a*f(x) dilates graph parallel to y-axis; scale factor 1/a. Similarly f(ax) dilates graph parallel to x-axis scale factor 1/a (since you need 1 a^(th) the same value of x to get the original f(x) so the graph must shrink horizontally) —Preceding unsigned comment added by Topology Expert (talk • contribs) 13:03, 17 September 2008 (UTC)

differential equation

Hi, is there a solution to xy" + y' - y = 0 that uses a finite set of well known functions?(Not the power series solution--I got the d.e. as something a certain power series satisfied and I want to express the function repped by the p.s. in terms of well known functions if possible. The function is y = the sum of (x^n)/[(n!)^2] from n=0 to infinity, which I got from the linear operator that sends the p.s. for 1/(1-x) to the one for e^x. When you do the same on the p.s. for e^x, you get y.)Thanks in advance.Rich (talk) 02:54, 17 September 2008 (UTC)

See Bessel-Clifford function with n=0. Dmcq (talk) 09:49, 17 September 2008 (UTC)

good referral, thank you.130.86.14.25 (talk) 05:10, 19 September 2008 (UTC)

combinations and permutaions

what is value of nC3 —Preceding unsigned comment added by 59.184.240.35 (talk) 13:13, 17 September 2008 (UTC)

See Combination. --Tango (talk) 13:19, 17 September 2008 (UTC)

Use the definition of binomial coefficient, substitute 3, and simplify. (Alternately, use the theorem that nCk is a polynomial of n of degree k for any fixed nonnegative integer k, compute it for a few small values, then interpolate. Alternately, compute for the first few values of n and look up in OEIS for a formula. Alternately rewrite the binomial to a falling power and use the equation about falling powers and Stirling numbers to get the result in expanded forms rightaway.)

If you don't know what binomial coefficients are in general then a good introduction might be the first few chapters of Graham–Knuth–Patashnik, Concrete mathematics. – b_jonas 15:15, 19 September 2008 (UTC)

September 18

Formula for planet temperature

I'm having trouble converting this formula from "FUNDAMENTALS OF ASTRONOMY" page 318 (QB43.3.B37 2006) by Dr.Cesare Barbieri, professor of Astronomy at the university of Pauda, Italy. It is not simplified for the Sun/Earth, it is the full formula with data for Gliese 581/581 c. Can someone help?

=(((((0.0000000567051)*(3840^4))/(4*PI()*((0.0613*149597876600)^2))) * ((4*PI()*((0.29*695500000)^2))/(4*PI()*((11162)^2)))*((PI()*((11162)^2)*(1-0.64))))/0.0000000567051)^0.25

$T_{p}={\frac {(5.67051E-8)*(3840^{4})}{(4*pi*(0.0613*1AU)^{2})}}*\ {{4*pi*(0.29*{R_{\odot }})^{2}}{(4*pi*(R)^{2}}}{pi*(R)^{2}*(1-0.296){}0.0000000567051)^{0}.25}$

--GabrielVelasquez (talk) 04:09, 18 September 2008 (UTC)

I think you will have to provide a few more details here. You have written two different expressions - which one (if either) is correct ? And in what way are you trying to "convert" the formula - what are you converying from and to ? Gandalf61 (talk) 08:59, 18 September 2008 (UTC)

I found the reference on Google books. You appear to be trying to use the second equation on the page, which looks like this (rewriting quantities that refer to the Sun to refer to a general star e.g R_⊙ becomes R_∗ and marking those which refer to the planet explicitly, e.g. T becomes T_p):

$\sigma \cdot T_{\mathrm {p} }^{4}={\frac {\sigma \cdot T_{\ast }^{4}}{4\pi \cdot d^{2}}}\cdot {\frac {4\pi \cdot R_{\ast }^{2}}{4\pi \cdot R_{\mathrm {p} }^{2}}}\cdot \pi \cdot R_{\mathrm {p} }^{2}(1-A_{\mathrm {p} })$

However I don't see why you are trying to use this equation, as many of the terms in it cancel each other out, making it unnecessarily cumbersome to work with. In fact it is rearranged on the next line as follows:

$T_{\mathrm {p} }=T_{\ast }\left[\left({\frac {R_{\ast }}{d}}\right)^{2}{\frac {1-A_{\mathrm {p} }}{4}}\right]^{\frac {1}{4}}$

This expression is far easier to work with (note for example that the radius of the planet cancels out) and gives exactly the same results! If you want to try it for yourself, I recommend Google calculator, which helpfully includes various units conversions and the constant r_sun, so for a planet orbiting, say, 0.1 AU from a 3500 K star with a radius 0.3 times that of the Sun and a bolometric albedo of 0.5, you could do a Google search for: 3500 K*((0.3*r_sun/0.1 AU)^2*(1-0.5)/4)^(1/4). Icalanise (talk) 12:53, 18 September 2008 (UTC)

Both of those formulas are in the book I mentioned, I just had trouble with the second divided terms in the code, thanks. I want the full version because I have seen other scientists use a different formula that looks wrong in comparison, but I know Selsis et al at Gliese 581 c simplifed it to factor out the Luminosity variables (L=4·pi·R^2·a·T^4)and it just looks wrong if you don't have the full version. GabrielVelasquez (talk) 21:16, 20 September 2008 (UTC)

Finding an average flood height overtime

dates of floods                   height of each flood
1965..........................    5.22
1972..........................    5.14
1993.........................     6.31
1997...........................   6.13

From this information iam required to find the average height of these floods for a 1 in 10 year event and a 1 in 100 year event

please help thankyou

if you need to email me it <email removed by Ζρς ι'β' ^¡hábleme!> the 0 is a zero —Preceding unsigned comment added by 124.170.38.135 (talk) 04:51, 18 September 2008 (UTC)

To find the average of values, just add them up, and divide by the number there are. For example, the average of 1, 2, 3, 4, and 5 is 3 because 1+2+3+4+5=15 and 15/5=3. Ζρς ι'β' ^¡hábleme! 21:53, 18 September 2008 (UTC)

Probably the best thing to do for real world problems is ask someone who deals with this in practice. Both the cases of flooding by the sea and of inland areas are quite complicated. You really don't have enough figures to come to any sort of even inaccurate conclusion without some model to plug them into. If you just want to use flood heights with nothing else you'd need many more figures from lower level floods so you can fit a probability distribution to them. Your figures above would be some way along one side and highly skewed and there's too few to see how quickly they go down. And anyway Global warming will make a mess of any figure just derived from historical data. Personally I'm rather surprised at the difference between 1972 and 1993 - I'd guess someone changed the environment of the area between the two. Dmcq (talk) 18:28, 20 September 2008 (UTC)

Even if the trend is towards higher floods, there is still going to be random fluctuations from year to year, it's not at all surprising that a later flood was a little lower than an earlier one. --Tango (talk) 19:51, 20 September 2008 (UTC)

Word for the "terms" in a union

Is there a word along the lines of "summand", "integrand", "radicand" for the individual "terms" in a union? For example, in $A\cup B\cup C$, what word describes the role of $A$, $B$, and $C$? Is "term" acceptable, or is there a niftier word like "unand" or "unitand"? While we're at it, what about intersections? —Bkell (talk) 05:01, 18 September 2008 (UTC)

I think terms in acceptable. As far as I know, there is no particular word associated to this. Thenub314 (talk) 07:31, 18 September 2008 (UTC)

I also have never heard such a specific term. If you wanted to invent one, then some people call union set summation, so you could call them summands. Or you could use the analogy with the logical OR operator and call them disjuncts. Algebraist 11:00, 18 September 2008 (UTC)

0 dimensional object in 4 dimensional world

In a tesseract, if the cube (3d) serves the function analogous to a 2d surface for a cube, and the plane serves the function analogous to the edge of a cube in 3d, what would be the status of a point(0 dimensional) in regard to four dimensions? It couldn't be the -1st dimension, could it?Leif edling (talk) 07:03, 18 September 2008 (UTC)

The analogy breaks down, or was never really more then an analogy. In the 4 dimensional cube you gain one more thing then you had before, you don't just translate all the things from a 3 dimensional cube. So in a 3 dimensional cube you have squares (2 dimensional cubes), line segments (1 dimensional cubes, this is a bit degenerate), and points (0 dimensional cubes, very degenerate). In a 4 dimensional cube there are, 3 dimensional cubes, squares, line segments and points. Thenub314 (talk) 07:40, 18 September 2008 (UTC)

Yeah, because if you stepped into 4d land, you'd have almost no mass and no weight, because you are infinitesimal thin. See Paper mario that would be you. Sentriclecub (talk) 11:48, 18 September 2008 (UTC)

My second answer, is just like a point in 3d. A point is so thin, so small, that even though I can look on a map and see a point 37 degrees noth, 48 degrees east. If I go to that actual location in the world, I still would not be able to even begin to see the "point" which exists, its smaller even the planck lengths. So my answer is that its just the same as in a 3d world. Remember, a point has no volume, so as soon as you give it a relative size, you instantly defined a cube (which is a 3d structure, which has volume). Sentriclecub (talk) 11:53, 18 September 2008 (UTC) My third answer is that it is a line, because given my second answer, that would say that a cube in 3d is not analogous to a tesseract (as you actually asked--I failed your original question) because a cube can exist in 4d just as a sheet of paper (to me at least) represents a 2D plane in my 3d world.

I read this "analogy" perspective:

Cells, Ridges, Edges: The upshot of all this is that in 4D, objects have a much richer structure than in 3D. In 3D, a polyhedron like the cube has vertices, edges, and faces, and fill a 3D volume. The cube is bounded by faces, which are 2D. Every pair of faces meet at an edge, which is 1D, and edges meet at vertices, which are 0D.

In 4D, objects like the hypercube not only has vertices, edges, and faces, but also cells. A 2D boundary is insufficient to bound a 4D object. Instead, 4D objects are bounded by 3D cells. Each pair of cells meet not at edges, but at 2D faces, also called ridges. The ridges themselves meet at edges, and edges meet at vertices.

The point here is that in 4D, 3D volumes play the role analogous to surfaces in 3D, and 2D ridges play the role analogous to edges. Because of this, it is important to visualize 4D objects by thinking in terms of bounding volumes, and not 2D surfaces. A 2D surface only covers the equivalent area of a thin string in 4D! When you see a 2D surface in the projection of a 4D image, you should understand that it is only a ridge, and not a bounding surface.

I got it here:[1]

So, do the above answers signify that a tesseract does not have 2d planes analogous to the edges of 3d cube? (As is said in the last paragraph above)Leif edling (talk) 15:49, 18 September 2008 (UTC)

Elliptic Curves Isomorphism

What is an isomorphism when it comes to elliptic curves? I have an idea of what it is based on reading about it in two different books, but neither seems very clear to me.

Knapp's book "Elliptic Curves", which I don't have with me so I can't tell you exactly what it says, basically says two elliptic curves are isomorphic if they are related by an admissible change of variables. That's the definition of isomorphic. But, it never says anything about what that means. I assume it means the group structures are the same. But, is that all it means?

The other book I looked at is Dale Husemoller's "Elliptic Curves". Again, I don't have it with me, but I believe one theorem says two elliptic curves are isomorphic if and only if they have the same j-invariant. This is why I am not entirely sure on the isomorphism meaning only group structure. If this is what it means, then since there are an infinite number of j-invariants, there must be an infinite number of elliptic curve groups possible. The problem is Knapp's book doesn't seem very clear on this. It says only that elliptic curves for ranks up to 12 are known but a fact like there are an infinite number of such groups means clearly elliptic curves of much higher rank exist (higher than any given number).

Can any one help me understand this better? Thanks. StatisticsMan (talk) 13:46, 18 September 2008 (UTC)

You seem to be a bit confused about what base field (or whatever) you are working over. If one has a change of variables making two curves isomorphic, the group structure will be the same, sure. But if I understand your question correctly, when you say "the group structures are the same" this means that points over some field form groups which are isomorphic. But this is not enough to say that the curves are necessarily isomorphic. If one looks at things over the complex numbers, all elliptic curve groups are isomorphic as abstract groups, or even as real Lie groups, being tori, the product of two circle groups. (But they aren't isomorphic as complex Lie groups / Riemann surfaces unless they have the same j-invariant.)

When Knapp? says that elliptic curves with ranks up to 12 are known, he is talking about the Q-rational points on the curve as a finitely generated abelian group (cf Mordell–Weil theorem). Again, this fga group does not characterize the curve. The infinitude of different groups doesn't follow from the infinitude of j-invariants, as many curves could have the same group of Q-rational points. A theorem of Barry Mazur shows that there are only finitely many possible torsion subgroups of the Q-points, so the possible infinitude of different groups must come from there being (possibly) infinitely many possible ranks, but as Knapp says, only ranks up to 12 are known.John Z (talk) 01:34, 19 September 2008 (UTC)

Well, you are correct that I meant over the rationals, so I apologize for not saying that. But, I still do not think I understand what an isomorphism for elliptic curves over Q really means. You told me that it's not just the group. Can you please tell me what it is since the books I have read don't bother to explain this? My question was based on Mazur's theorem which says there are only 15 possible torsion subgroups of Q-points. So, since I do not understand what an isomorphism means, I made a guess that it meant only the group. If it means only the group, it follows there must be an infinite number of different ranks. But, Knapp says only up to 12 are known. So, it's either not just the group or it is and his statement does not say all it should. StatisticsMan (talk) 15:24, 19 September 2008 (UTC)

I'm not familiar with this stuff at all, but it probably means they're isomorphic as algebraic varieties. Algebraist 22:03, 19 September 2008 (UTC)

Yes, that is all that it is, an invertible morphism of varieties. In StatisticsMan's words "related by an admissible change of variables" - we just need to change variables in an invertible way. An elliptic curve may be given by an equation like y^2=x^3+ax^2+bx+c. If we did a simple invertible change of variable, say replacing x by x+1, we get an equation of the same type that thus gives an isomorphic curve embedded trivially differently in the plane. The j-invariant can be defined algebraically from the coefficients (a,b,c) of such an equation and is invariant under any such coordinate change. An isomorphism of the groups of rational points of two different curves with different j-invariants is not necessarily induced by an algebraic change of coordinates relating the two curves. With elliptic curves it is practical to see modern algebraic geometry at work in a very concrete fashion. For this one might want to look at the lucid paper of John Tate, The Arithmetic of Elliptic Curves, Inventiones Math.(23) 1974, 179-206, which later expository works on elliptic curves owe much to. Things like Mazur's theorem are much harder.John Z (talk) 06:10, 20 September 2008 (UTC)

So, just to be clear, we are saying that curves with isomorphic groups may have different j-invariants, so they are not isomorphic as curves, even thought their groups are isomorphic. So, although it is conjectured that there is no upper limit on the possible rank of a curve group, the fact that we can construct curves with an infinite number of different j-invariants does not prove this conjecture.

Incidentally, the search for curves whose groups have high rank has moved on a bit - this page says that the highest rank of an elliptic curve whose group's rank is known exactly is 18, and the highest known lower limit on a curve's group's rank is 28 (i.e. there is a curve whose group's rank is known to be at least 28, but it may be higher). Both examples were found by Noam Elkies in 2006. Gandalf61 (talk) 09:04, 20 September 2008 (UTC)

Alright, thanks a lot everyone. This has been very helpful. I appreciate that you took the time to answer my question.StatisticsMan (talk) 15:15, 20 September 2008 (UTC)

Is there any limitation to defining all x in Q as...

$x={\frac {m}{n}}$, where $n,m\in \mathbb {Z}$ and $n>0\,$? I do not mean we need to show that Q is a field, but I have seen other "loose" definitions such as $n,m\in \mathbb {Z} ,n\not =0$ and I just thought that having n and m as negative was redundant, since we only need the numerator, m to be in Z.

Yes, a common way of doing this is defining $\mathbb {Q}$ as the set of equivalence classes of these formal fractions ${\frac {m}{n}}$, defining two fractions as equal if $m_{1}n_{2}=m_{2}n_{1}$. When doing this, it does not matter at all if you require $n$ to be positive or just non-zero. The reason for choosing just non-zero is probably that then the same process will work for non-ordered domains. In other words: for constructing rationals from integers, your approach is just as good; but it's less adaptable. -- Jao (talk) 16:30, 18 September 2008 (UTC)

One use for the signs of the integers in computational work is to specify that $(-a)/b$ and $a/b$ (ie not $a/(-b)$ and $(-a)/(-b)$) mean that the fraction is not known to be in its lowest terms. It's a nice convenient flag that means "you do not need to try and reduce this fraction" Robinh (talk) 07:03, 19 September 2008 (UTC)

Algebra

What are some applications of groups, either to other mathematical structures or to the real world? Other than a little bit of Galois theory and a mention of homology, my textbooks haven't given much indication of what it's for. Black Carrot (talk) 22:49, 18 September 2008 (UTC)

Quantum chromodynamics, The Standard Model, and related areas of particle physics are littered with group theory. Dragons flight (talk) 22:52, 18 September 2008 (UTC)

Group theory is used extensively in chemistry, too. I don't know very much chemistry, but I believe groups are used in crystallography at least. Perhaps a chemist will come along and can provide more information. Another critically important application of group theory is the analysis of the Rubik's Cube. ;-) —Bkell (talk) 04:29, 19 September 2008 (UTC)

Correct about chemistry/crystallography. Spectroscopy as well. The types of events that give rise to spectroscopic signals are determined by group theoretical properties of the molecule's symmetry. See the book mentioned here for several such applications. Baccyak4H (Yak!) 16:20, 19 September 2008 (UTC)

Groups have lots of applications in functional analysis, harmonic analysis and algebraic topology (although vector spaces are also used).

Topology Expert (talk) 05:51, 20 September 2008 (UTC)

Elliptic curve cryptography. Gandalf61 (talk) 08:51, 19 September 2008 (UTC)

September 19

probability

what is the probability of 1.0 means? and between probability of 1.0 and o.4 which the best to present the survival without parasitemia —Preceding unsigned comment added by 41.221.34.123 (talk) 07:53, 19 September 2008 (UTC)

A probability of 1 means that it is certain (100% likely) to occur. A probability of 0 means it is certain to never occur, 0.5 means the event will happen half (50%) of the time. A probability cannot be less than 0 nor can it be greater than 1. -- SGBailey (talk) 07:58, 19 September 2008 (UTC)

This is wrong. If an event has probability 1, it does not mean that it is certain. For example, consider the event of randomly picking a number in [0,1]. The probability that you pick a number other than 0 is 1 but it is not certain that you will not pick a 0.

See Almost surely.

Topology Expert (talk) 05:44, 20 September 2020 (UTC)

Please clarify your second question. Are you asking whether 1 or 0.4 is a more reasonable measure of a survival rate without parasitemia? Note that survival rates are usually associated with a certain length of time. Zain Ebrahim (talk) 14:15, 19 September 2008 (UTC)

Though maybe it should be noted, from a mathematical and not at all practical stand point, that probability 1 is not quite certain and probability 0 is not quite "certainly not". Thenub314 (talk) 14:57, 19 September 2008 (UTC)

Eh? From a mathematical point of view, 1 is certain and 0 is certainly not. From a colloquial point of view then anything goes. Use "Acme's gadget for 110% success" is gibberish mathematically. -- SGBailey (talk) 15:16, 19 September 2008 (UTC)

1 is not certain; see my previous comment.

Topology Expert (talk) 05:47, 20 September 2008 (UTC)

What Thenub314 means is that any set with measure zero has probability zero, not just the empty set. Pick a random real number between 0 and 1. The probability of your picking that number was 0, yet you picked it. -- BenRG (talk) 16:28, 19 September 2008 (UTC)

We have an article: almost surely. Algebraist 16:31, 19 September 2008 (UTC)

Note that there is a subtle difference here between the theory and practice in that we have no way to pick a random real number between 0 and 1. So when you pick a random real number, it actually has some nonzero probably. Even with a true random number generator, there is only a finite list of expressible numbers.

To my knowledge, everything in practice is a finite process, and so there is no practical distinction between almost surely, and certain: and the OP probably doesn’t know any measure theory. GromXXVII (talk) 13:30, 21 September 2008 (UTC)

September 20

Planes and Spheres

I need a bit of help understanding a problem I'm doing and Wikipedia seemed like the best place.

A plane $P_{1}$ has equation

$3x-y-1=0$

and a sphere $S_{1}$ has equation

$x^{2}+y^{2}+z^{2}=7$.

I have to determine the equation of the circle, $C_{1}$, defined by the intersection of $P_{1}$ and $S_{1}$.

I have tried this in a number of ways and nothing seems to help. From the equation for $P_{1}$, I can see that $y=3x-1$. Substituting this into the equation for $S_{1}$ gives $x^{2}+(3x-1)^{2}+z^{2}=7$, which is simplified to

$10(x-{\frac {3}{10}})^{2}+z^{2}={\frac {69}{10}}$

This clearly isn't a circle and plotting it in my graphing application confirms it.

My next method was to rearrange the equation for $P_{1}$ to arrive at $21x-7y=7$. Substituting this into the equation for $S_{1}$ gives $x^{2}+y^{2}+z^{2}=21x-7y$, which eventually simplifies to

$(x-{\frac {21}{2}})^{2}+(y+{\frac {7}{2}})^{2}+z^{2}={\frac {490}{4}}\,\!$.

This is even more confusing because somehow the intersection of plane and a sphere has become a sphere, defying all logic. Could someone please tell why these two methods have failed and point me in the right direction, ie don't explicitly tell me what to do, just give a hint? I want to try and do as much of this by myself as possible. Thanks 92.2.212.113 (talk) 15:22, 20 September 2008 (UTC)

A circle, a 1-dimensional thing, cannot be represented by 1 equation in 3-D, because each equation only constrains one dimension; so you will need 2 equations to represent your circle. --71.147.13.131 (talk) 17:17, 20 September 2008 (UTC)

Indeed. Baring degenerate cases, each equation (or "constraint") reduces the dimension by one. You've started in 3D space, so one constraint gives you a 2D subset, another gives you 1D and a third would give you a single point (0D). You can't really get much better than just giving the equations of the sphere and the plane - the question seems to be flawed. Perhaps it wants the equation of the projection of the circle on to one of the co-ordinate planes? --Tango (talk) 19:49, 20 September 2008 (UTC)

Do the two constraints come from the fact that the equation for the plane involves both x and y? 92.2.212.113 (talk) 19:54, 20 September 2008 (UTC)

Perhaps clarity will be aided with the complete question. $P_{1},S_{1},C_{1}$ are the same as above. $P_{2}$ is the plane with equation $x-y+1=0$, $S_{2}$ is the sphere with equation $x^{2}+y^{2}+z^{2}-6y-4z+10=0$, $C_{2}$ represents their intersection and L is the intersection of $P_{1}$ and $P_{2}$. "Find the points where L meets each of $S_{1}$ and $S_{2}$. Determine, giving your reasons, whether the circles $C_{1}$ and $C_{2}$ are linked." Does that help? 92.2.212.113 (talk) 20:02, 20 September 2008 (UTC)

Since a circle has an area, I see no way that it can be a 1-dimensional object. Also, if a plane intersects a sphere, not if it just touches the sphere, the points of intersection must produce a circle, see here Plane-sphere intersection. 92.2.212.113 (talk) 19:50, 20 September 2008 (UTC)

Hmm I think my confusion stems from thinking that the intersection produce a disc as well as a circumference. It's now clear that it only produces a circumference. This question comes from a well established exam though, so there must be something I've missed out that would make it work. 92.2.212.113 (talk) 19:53, 20 September 2008 (UTC)

The two constraints are the two equations, one for the plane and one for the sphere, each equation will be one constraint, regardless of how many variables are in it (as long as it isn't a vector equation, anyway - vector equations represent multiple scalar equations). I don't see anywhere in that question that asks for an equation for the circle, you just need to determine if the two circles are linked, that doesn't require having an equation for the circle. --Tango (talk) 20:42, 20 September 2008 (UTC)

OK, I'm with you on the constraint issue. So is it impossible to determine an equation for the circles then? If so, I suppose a new method of approach is called for... Thanks 92.2.212.113 (talk) 21:26, 20 September 2008 (UTC)

It is impossible to determine a single equation for the circle (barring messy stuff), but it is possible to find a pair of equations for the circle: there are many such pairs, one such being the pair of equations you started with, as Tango pointed out. Try approaching your problem from some other direction (and ask for assistance if wanted). Eric. 65.96.172.100 (talk) 22:20, 20 September 2008 (UTC)

The "messy stuff" doesn't have to be all that messy: $(3x-y-1)^{2}+(x^{2}+y^{2}+z^{2}-7)^{2}=0$ would do it. Of course, that is absolutely no help in solving the main problem ;) AndrewWTaylor (talk) 13:26, 22 September 2008 (UTC)

The main point of this question seems to be to provoke the answerer into providing a proper definition of that mysterious word "linked". Once you've done that, it should be obvious how to proceed with the equations. (My definition wouldn't involve the line L at all, and my method wouldn't really make use of those line/sphere intersection points, but I can see how they're related.) --tcsetattr (talk / contribs) 22:00, 20 September 2008 (UTC)

Just a note: A circle is a 1-manifold and therefore is one-dimensional in the topological sense.

Topology Expert (talk) 13:45, 21 September 2008 (UTC)

Radius vs. radian

The mean circumference of Earth is about 40041.5 km, which means 1° of arc ≈ 111.2263 km, or 1 rad of arc ≈ 6372.8 km, as 1 rad = 57.295779513...°. A radian ("rad") is the arc equivalent of radius. So why is the inverse of curvature called "the radius of curvature", and not "a radian of curvature"? Since 1° of arc, or "arcdegree", is 1/360° of the circumference, isn't 1/(2π), here 6372.8, the "arcradian", rather than "arcradius".  ~Kaimbridge~ (talk) 17:06, 20 September 2008 (UTC)

The radius of curvature is so called because it is the radius of a circle with the same curvature as the given curve. It would be silly to say that a curve has x radians of curvature rather than radius of curvature x for at least two reasons. Firstly, radius of curvature is a dimensional quantity, in fact a length. Do you want to say that the surface of the earth has 6372.8 km radians of curvature? Secondly, the 'of curvature' suggests that a higher number corresponds to higher curvature, which is of course not the case. Algebraist 17:11, 20 September 2008 (UTC)

Of course not plural, anymore than you would say Earth has "6372.8 km radii of curvature"!——one radian is the arc equivalent length of a radius: In a unit circle the radius = 1, a radian = 1 and the circumferemnce = 6.2831853...radians. So if a circle's radius = 6372.8, then one radian = 6372.8 and the circumference = 40041.5. Thus Earth would have a radian of curvature equaling 6372.8. See Plane and Spherical Trigonometryand Elements of Geometry.  ~Kaimbridge~ (talk) 22:47, 20 September 2008 (UTC)

What do you mean by "a radian = 1"? A radian is a radian, it doesn't equal a number. --Tango (talk) 22:54, 20 September 2008 (UTC)

See the definition of radian:

"One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle."  ~Kaimbridge~ (talk) 01:25, 21 September 2008 (UTC)

Yes, I know what a radian is, but it's a measure of angle, it isn't equal to a natural number. --Tango (talk) 13:23, 21 September 2008 (UTC)

Is it accurate to use "1° of arc" like that? I thought the "of arc" was just to distinguish between other things measured in degrees (temperature, for example). "1° of arc" is still a measure of angle, not length, by my understanding. --Tango (talk) 19:44, 20 September 2008 (UTC)

Yes, see the links I gave above, and this application of it(73.208(c)(3)-(4)):

(3) Calculate the number of kilometers per degree latitude difference for the middle latitude calculated in paragraph (c)(2) as follows:

KPD[lat]=111.13209−0.56605 cos(2ML)+0.00120 cos(4ML)

(4) Calculate the number of kilometers per degree longitude difference for the middle latitude calculated in paragraph (c)(2) as follows:

KPD[lon]=111.41513 cos(ML)−0.09455 cos(3ML)+0.00012 cos(5ML)

Here, KPD[lat] is "M" and KPD[lon] is "N(×cos(ML)", the principal radii of curvature in degree form of their series expansion, and the latitude and longitude differences are angular distances (i.e., "degrees of arc").  ~Kaimbridge~ (talk) 22:47, 20 September 2008 (UTC)

That quote doesn't include the phrase "degree of arc", so how does it answer my question? --Tango (talk) 22:54, 20 September 2008 (UTC)

It says "per degree latitude difference" and "per degree longitude difference". As I said above, the latitude and longitude differences are "degrees of arc", so one degree of latitude difference equals one degree of arc.

Let's say ΔLat = 20°: Using the previously cited mean values, distance = 2224.526 km = 20°× 111.2263 = .3490658 rad × 6372.8. What would you call 111.2263 and 6372.8? I would say 111.2263 is the value of an arcdegree and 6372.8 is the arcradius, though it would seem it should be arcradian.  ~Kaimbridge~ (talk) 01:25, 21 September 2008 (UTC)

I've never seen the terms used that way before, although I now see Nautical mile uses them that way. I would have just said "the arc length corresponding to one degree" or something. Even with those definitions, "radian of curvature" wouldn't make sense, a radian is still a measure of angle, you could say "radian of arc of curvature" since a radian of arc is, by definition, the radius, but I don't see much point. --Tango (talk) 11:15, 21 September 2008 (UTC)

Okay, then that brings the question——he-he——full circle: Should it be "arcradius" or "arcradian".

Consider example 2, here:

Example 2. An angle of .75 radians means that the arc is three fourths of the radius. s = .75r

So letting r = 6372.8, we have

4779.6 =.75·6372.8 = 42.9719°·111.2263,

thus an arcdegree = 111.2263, but what is 6372.8, the arcradius or an arcradian? Googling around, I find several references to arcradius, though mostly referencing height and width calculations of circles, but none for arcradian (or "arc radian"), with regard to what I'm talking about. On one hand, it would seem arcradian belongs with arcdegree, arcminute and arcsecond, not arcradius, but on the other we are talking about the arc equivalent of the plane radius, which would suggest arcradius.  ~Kaimbridge~ (talk) 17:48, 21 September 2008 (UTC)

Using your definitions, "arc-radian" would mean the same as "radius". I have no idea what "arc-radius" would mean. --Tango (talk) 18:01, 21 September 2008 (UTC)

Integration

sin(Tan^-1) is not a function? —Preceding unsigned comment added by Saadmunir (talk • contribs) 20:22, 20 September 2008 (UTC)

Not in the common sense, no. It's a multivalued function though, with all input values (except 0) having two output values. (I don't see what this has to do with integration.) -- Jao (talk) 20:32, 20 September 2008 (UTC)

If you restict the inverse tan to the principal branch, as is commonly done, it's fine, though. --Tango (talk) 20:34, 20 September 2008 (UTC)

Isn’t the symbol tan^-1(x) by definition only defined on $({\frac {\pi }{2}},{\frac {\pi }{2}})$? GromXXVII (talk) 13:07, 21 September 2008 (UTC)

It's often (although not universally) defined to take values only on that interval, is that what you mean? It's defined everywhere. --Tango (talk) 13:22, 21 September 2008 (UTC)

Err yeah that’s what I meant. GromXXVII (talk) 13:35, 21 September 2008 (UTC)

Note also that sin(tan⁻¹(x)) = x/√(1 + x²). Michael Hardy (talk) 05:09, 21 September 2008 (UTC)

functions

what is parent function and daughter function? —Preceding unsigned comment added by Saadmunir (talk • contribs) 20:24, 20 September 2008 (UTC)

Looking at Google I see parent function is as I suspected yet another of those ?#@[+! terms invented by people in education. This one meaning basic functions like exp or sin like those on a calculator. I didn't see anything about daughter function. Aged parents often believe it is to look after them in their dotage but daughters tend to have other ideas about it. I can't see where the maths comes into that though. :) Dmcq (talk) 21:31, 20 September 2008 (UTC)

Child function, perhaps? (Being gender-sensitive). I did find [2] which is a horrible way of thinking about things, I think. I believe "parent functions" refer to a "general" function, such as $f(x)=x^{2}$, while "child functions" refer to thinks like $f(x+1)=(x+1)^{2}$ or $f(x+3)=(x+3)^{2}$, where f is defined previously. The closest I learnt to this concept would be "family of curves". I think it mostly refers to translation although I can't find any definitions, so to restrict this to translation would probably be wrong. x42bn6 Talk Mess 21:01, 21 September 2008 (UTC)

A family of mathematical objects is usually a collection of objects where each is on an equal footing, I wouldn't distinguish between children and parents. --Tango (talk) 21:06, 21 September 2008 (UTC)

Algebra

-6 = -6

---

9-15 = 4-10

---

9-15+ 25/4 = 4-10+ 25/4

---

(3-5/2)² = (2-5/2)²

---

3-5/2 = 2-5/2

---

3 = 2

---

what's wrong in it? —Preceding unsigned comment added by Saadmunir (talk • contribs) 20:34, 20 September 2008 (UTC)

When you take the square-root, there are two possibilities. You can have them both with the same sign, as you've done, or you can have them with different signs, which gives 3-5/2=-2+5/2, which is true. See Extraneous and missing solutions. --Tango (talk) 20:38, 20 September 2008 (UTC)

More directly, ${\sqrt {x^{2}}}=|x|$. Dragons flight (talk) 21:22, 20 September 2008 (UTC)

Or, just look at line four, then simplify it *before* taking the square root: (1/2)² = (-1/2)². From there, if you can conclude that +1/2 = -1/2, then maybe 3 does equal 2! --DaHorsesMouth (talk) 22:51, 20 September 2008 (UTC)

It all boils down to this:

Even if x^2 = y^2

you CANNOT conclude that x = y (without additional information or additional constraints)

122.107.140.79 (talk) —Preceding undated comment was added at 00:57, 21 September 2008 (UTC).

See also Invalid proof. This is one of the classic tricks in deliberately invalid proofs. PrimeHunter (talk) 01:30, 21 September 2008 (UTC)

September 21

Puzzle

I heard this a little while ago, and I don't think I've asked it here yet. I don't know the answer, and so far as I know neither does the guy who came up with it. Given a simple graph, choose a well-ordering of the points. Following that ordering, label each point with an ordinal number, subject to the following restriction: The label must be as small as possible, without being equal to the label of any adjacent point that's earlier in the ordering. The Grundy number of the graph is the supremum, in the ordinals, of the labels of all the graph's points, over all possible well-orderings of the graph. So for instance, the graph on the points A, B, C, and D with the lines AB, BC, and CD would have Grundy number 2, since that's the highest number you get in the graph from any ordering of its vertices. The possible labelings are 0101, 0120, 1010, and 0210. The question is, what's the Grundy number of the plane, where two points are adjacent if they're unit distance apart? I can get it up to the first uncountable ordinal, using a bit of recursion and some perturbation arguments, because I can produce any countable ordinal as a label of the graph, but I don't know if it's possible to specify an ordering that would produce an uncountable ordinal, or where you would go from there. Any ideas? Black Carrot (talk) 22:57, 21 September 2008 (UTC)

A related question, that may be easier: Will it be a limit ordinal? Black Carrot (talk) 22:58, 21 September 2008 (UTC)

Elliptic curves with isomorphic group structure that are not isomorphic

This is a continuation of a question I asked a few days ago. If two elliptic curves are isomorphic, it means more than just that their group structures are isomorphic. So, are there easy examples of two elliptic curves over the rationals that are not isomorphic but their group structures are known and are isomorphic? StatisticsMan (talk) 23:53, 21 September 2008 (UTC)

Consider

$y^{2}=x^{3}+x\,;\,j=1728\,,$

$y^{2}=x^{3}-1\,;\,j=0\,.$

Curves are not isomorphic as curves, because they have different values of j. But for both curves, rank is 0 and group of rational points is isomorphic to C₂; the only rational points on these curves are (0,0) for the first curve and (1,0) for the second (each curve group also includes a point at infinity, which is its identity element). Gandalf61 (talk) 10:52, 22 September 2008 (UTC)

Thank you once again. That is a very easy example. 129.186.52.59 (talk) 15:41, 22 September 2008 (UTC)

September 22

Dedekind cuts

From the article:

A typical Dedekind cut of the rational numbers is given by

$A=\{a\in \mathbb {Q} :a^{2}<2\lor a\leq 0\},$

$B=\{b\in \mathbb {Q} :b^{2}\geq 2\land b>0\}.$

This cut represents the irrational number ${\sqrt {2}}$ in Dedekind's construction.

Would one need to use both sets to represent ${\sqrt {2}}$, or just one? Defining it as A would imply the existence of B, and vice versa; i.e. if I wanted to define ${\sqrt {2}}$, I would just have to describe A or B, but is there any need for both outside of being totally rigorous? —Preceding unsigned comment added by 81.187.252.174 (talk) 07:23, 22 September 2008 (UTC)

See Dedekind_cut#Handling_Dedekind_cuts: "each of A and B does determine the other. It can be a simplification, in terms of notation if nothing more, to concentrate on one 'half' ". Bo Jacoby (talk) 09:30, 22 September 2008 (UTC).

Find the natural domain and correspong range of this function...

Note: if someone can and doesn't mind doing so, please fomat the notation properly so that it is easier to read.

$x\rightarrow {\sqrt {x(x-3)}}$

Im not really interested in the working or anything. I just want to know if someone can verify that:

• the domain is the set real numbers "greater than or equal to 3" or "greater than or equal to 0"
• the range is the set of real numbers "less than or equal to 0".

Thats what the book and my teacher say. I agree with the domain because you cant end up with a negative inside the square root. But I dont agree with the range because the square root of a number can be positive or negative. Therefore the range should just be the set of real numbers.

--RMFan1 (talk) 13:08, 22 September 2008 (UTC)

The square root function returns only one result. Yes, it's true that there are numbers a and -a square roots for some c, but the square root function returns only the positive number. x42bn6 Talk Mess 13:16, 22 September 2008 (UTC)

Haven't you got the domain the wrong way round? x(x-3) is negative for x between 0 and 3, so the domain is *greater* than or equal to 3 or *less* then or equal to 0. --Tango (talk) 13:22, 22 September 2008 (UTC)

Also, your book should have specified that you are expected to find the domain and range on the basis that this is a function from real numbers to real numbers. If it is regarded as a function from complex numbers to complex numbers then its domain and range are the whole of the complex plane - in fact, it covers its range twice, with the exception of one critical value. (although this is sort-of-implied by the use of the square root function). Gandalf61 (talk) 13:38, 22 September 2008 (UTC)

Ok thanks. The error then is I, along with pretty much everyone else in the class take the square root function as returning the positive and negative values. Even the teacher so I dont know why she didnt think the book was wrong too --RMFan1 (talk) 13:51, 22 September 2008 (UTC)

Well, it depends on what you mean by "square root function". The square root symbol means "positive square root", that's almost universal, the words "square root", though, mean a multi-valued function, as does $x^{\frac {1}{2}}$. --Tango (talk) 14:26, 22 September 2008 (UTC)

I don't know what you were taught, but both "square root" and $x^{\frac {1}{2}}$ imply only the positive branch in my experience. The square root of 25 is 5, not +/- 5. Dragons flight (talk) 21:16, 22 September 2008 (UTC)

In my experience, x^1/2 is only required to be positive if x is implicited constrained to be a positive real. Otherwise we're using the complex exponential function, and it's inherently two-valued. Algebraist 21:19, 22 September 2008 (UTC)

So you'd say (-25)^1/2 is +/- 5i? I'd still say this is only +5i, i.e. the positive branch. What to do with complex x might be more ambiguous, but in my experience all real valued x would be understood as implying the positive branch. Dragons flight (talk) 21:29, 22 September 2008 (UTC)

What do you consider to be the positive branch of the complex square root function? In my experience, there's no clearcut answer, and at least one standard answer is undefined precisely on the negative reals. Algebraist 21:40, 22 September 2008 (UTC)

Rearranging equations

I have been given the function $f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}$ and am told that $f'(x)={\frac {(x-1)}{2x{\sqrt {x}}}}$.

I get that $f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}=x^{\frac {1}{2}}+x^{-{\frac {1}{2}}}$ and that $f'(x)={\frac {1}{2}}x^{-{\frac {1}{2}}}+-{\frac {1}{2}}x^{\frac {3}{2}}$ from this.

How do I go from here to the final answer of $f'(x)={\frac {(x-1)}{2x{\sqrt {x}}}}$? No matter what I do, I keep on getting stuck. Thanks. --AFairfax (talk) 16:58, 22 September 2008 (UTC)

Surely you mean $f'(x)={\frac {1}{2}}x^{-{\frac {1}{2}}}+-{\frac {1}{2}}x^{-{\frac {3}{2}}}$ (note the negative exponent on the second term)? -- Jao (talk) 17:03, 22 September 2008 (UTC)

(Assuming the missing minus sign was just a typo) Factorise out an $x^{\frac {-3}{2}}$ term. --Tango (talk) 17:14, 22 September 2008 (UTC)

Indeed it was, thanks for pointing that out. Still, I'm not getting anywhere factorising out an $x^{\frac {-3}{2}}$ term. I manage to get: $x^{-{\frac {3}{2}}}\left({\frac {1}{2}}x^{-{\frac {4}{2}}}+{\frac {1}{2}}x^{-{\frac {1}{2}}}\right)$ —Preceding unsigned comment added by AFairfax (talk • contribs) 20:20, 22 September 2008 (UTC)

I've no idea how you got that. Factor out the x ^-3/2 term correctly, and you have the given answer. Algebraist 21:00, 22 September 2008 (UTC)

I thought that might be the problem, but I have no idea how to do it correctly. Nor can I get the suggestion below to work (it's too late, and my Maths is too bad...) --AFairfax (talk) 21:35, 22 September 2008 (UTC)

To work it backwards, split the fraction into two parts: ${\frac {x}{2x{\sqrt {x}}}}-{\frac {1}{2x{\sqrt {x}}}}$. Then just simplify it. To work it forwards (which is always the better way to learn - you can't rely on always having the answer in front of you!), "factorise out by X" means "divide each term by X, put brackets around the whole thing, and write 'X' at the front". --Tango (talk) 21:54, 22 September 2008 (UTC)

I understand the concept of factorising, it's just that in this case, I don't appear to be able to factorise out x ^-3/2 correctly. Thanks for all the help guys. --AFairfax (talk) 22:34, 22 September 2008 (UTC)

It's easy to work backwards from the known answer. Just make the division in $f'(x)={\frac {(x-1)}{2x{\sqrt {x}}}}$ and compare. PrimeHunter (talk) 21:09, 22 September 2008 (UTC)

Following up on Tango's explanation of factorization, here we go step-by-step. We wish to factor $x^{-3/2}$ out of the two terms $x^{-1/2}-x^{-3/2}$ (I'll forget about multiplying by 1/2, it doesn't matter). First we divide each term by $x^{-3/2}$:

${\frac {x^{-1/2}}{x^{-3/2}}}=x^{-1/2}\cdot x^{3/2}=x^{-{\frac {1}{2}}+{\frac {3}{2}}}=x^{1}=x$

${\frac {-x^{-3/2}}{x^{-3/2}}}=-x^{-3/2}\cdot x^{3/2}=-x^{-{\frac {3}{2}}+{\frac {3}{2}}}=-x^{0}=-1$

Then we put brackets about and multiply back by $x^{-3/2}$:

$x^{-3/2}\cdot (x-1)$

Which is the same thing as:

${\frac {x-1}{x^{3/2}}}={\frac {x-1}{x{\sqrt {x}}}}$

Since we have first divided by $x^{-3/2}$ and then multiplied by $x^{-3/2}$, the resulting expression is equal to the expression we started with. I've written the steps out in some detail to help you find where you might have made a mistake in your own work. Eric. 65.96.172.100 (talk) 04:29, 23 September 2008 (UTC)

Mean of circular quantities

The article says:

"Convert all angles to corresponding points on the unit circle, e.g. α to (cosα,sinα). That is convert polar coordinates to Cartesian coordinates. Then compute the arithmetic mean of these points. The resulting point will lie in the unit circle.'"

Shouldn't it be unit disk? 83.23.221.221 (talk) 22:16, 22 September 2008 (UTC)

I think either "within the unit circle" or "on the unit disk" would work, "in the unit circle" is ambiguous. I've changed it to "on the unit disk", which seems the better of the two options to me (since a point on the unit circle might not be considered to be within, but it is on the (closed) unit disk). --Tango (talk) 22:28, 22 September 2008 (UTC)