Wikipedia:Reference desk/Archives/Mathematics/2010 March 20

= March 20 =

Integration by substitution
Hey, it's me again. I read the article about Trigonometric substitution but there is one part that it was very vague on. "For a definite integral, one must figure out how the bounds of integration change. For example, as x goes from 0 to a/2, then sin(θ) goes from 0 to 1/2, so θ goes from 0 to π/6. Then we have" How does one figure out how the bounds of integration change? Thanks, and sorry if this question is a lot simplr than the ones you usually get! ~ ~ ~ ~ —Preceding unsigned comment added by 76.230.225.102 (talk) 03:26, 20 March 2010 (UTC)
 * Well in this case you have x = a sinθ. The bounds 0 to a/2 indicate that we are starting at x = 0, and ending at x = a/2.  So if we want to integrate in terms of θ instead, we need to find what θ is in those places.  Using our formula for converting x into θ, we find that when x = 0 we have θ = 0 and when x = a/2 is when θ = π/6, so integrating from where x = 0 to x = a/2 is the same as integrating from where θ = 0 to θ = π/6. Rckrone (talk) 04:16, 20 March 2010 (UTC)
 * Using Iverson brackets makes things easier. Substituting x=g(u) where g is an increasing differentiable function, gives
 * $$\int_a^b f(x)dx$$ $$=\int [a<x][x<b]f(x)dx$$ $$=\int [a<g(u)][g(u)<b]f(g(u))d(g(u))$$ $$=\int_{g^{-1}(a)}^{g^{-1}(b)}f(g(u))g'(u)du$$. Bo Jacoby (talk) 16:34, 20 March 2010 (UTC).

We had
 * $$ x = a\sin\theta. \, $$

so as x goes from 0 to a/2, then a sin &theta; goes from 0 to a/2. Therefore sin &theta; goes from 0 to 1/2. You need to remember some trigonometry: sin 0 = 0 and sin(&pi;/6) = 1/2. Michael Hardy (talk) 16:52, 20 March 2010 (UTC)

Interest computation- when compounding frequency is lesser than pay out frequency
How should interest computation be done when the interest is required to be paid out earlier than the contracted pay out date considering compound interest. For e.g A fixed deposit is placed for 1 year for $10000 with 5% interest p.a, compounded monthly. The amount (principal+interest) is to be paid at the end of the term. Now if the interest so compounded was to be paid out quaterly, how should the discounting of interest be done? what is the formula for doing this calculation? —Preceding unsigned comment added by Motuammu (talk • contribs) 08:35, 20 March 2010 (UTC)
 * The accumulation after 3 months (a quarter) will be $$10,000(1+\frac{5\%}{12})^{3}$$ but I think we need more details. What do they pay out at each quarter? If it's the increase in the account, it will be the above number less 10,000. In that case, after they payout the account goes back to 10,000. Zain Ebrahim (talk) 10:11, 20 March 2010 (UTC)
 * Dividing 5% by twelve rather than taking the twelfth root of 1.05 is only an approximation, although we don't know exactly how the bank etc does its calculations. 92.29.149.119 (talk) 20:55, 21 March 2010 (UTC)
 * No, that's wrong. If the interest is compounded m times per period, then the annual rate per period compounded m times is the effective rate for $$\frac{1}{m}$$ periods multiplied by m. This is the definition of compounding multiple times per period. Zain Ebrahim (talk) 21:45, 21 March 2010 (UTC)
 * That was not how my credit-card debt used to be calculated each month. In any case, the exact method of calculation is going to vary from bank to bank - they are probably going to choose a method which works most in their favour. 84.13.47.185 (talk) 16:04, 22 March 2010 (UTC)
 * See my comment below. Zain Ebrahim (talk) 07:37, 23 March 2010 (UTC)
 * See mine below yours. 78.149.133.100 (talk) 22:13, 23 March 2010 (UTC)
 * I think the article Annual percentage rate will help the OP. The nominal APR, vs. the effective APR, represents the spread between equivalent interest rates that are either compounded or single-payout at end of term.  Also keep in mind that some financial contracts penalize pre-term withdrawal.  Nimur (talk) 11:37, 20 March 2010 (UTC)


 * Reading the OP's question above, you could use a monthly formula as the basis for both calculations. First calculate the monthly interest rate - it will be the twelfth root of 1.05, minus 1. I make it about 0.4 percent. So each month you get about an extra 0.4% of your principal added to it, making the principal slightly bigger. You use this revised principal in your calculations for next month. In the first case you do this (without withdrawing any interest) for twelve months. In the second case you do this for only three monthas, then withdraw all the interest, so you are back with your original principle, and repeat this procedure four times. Note that the total amount of interest you get in the second case will be less that what you would get at the end of the year in the first case, as you've spent the interest rather than leaving it in the account to grow, even though the interest rate of the two cases is the same.


 * If you are asking this question because you want to check if a bank is correct or not, then be aware that the details of how they do their calculations can vary considerably, and can be difficult to understand (continuous compounding for example). So as someone suggested above, best to use the APR of each account to compare them. APR is carefully standardised in the EU and UK, although in non-EU countries such as the USA that is not so true. Similarly the interest rate they tell you is usually substantially different from the APR. 84.13.41.17 (talk) 15:10, 21 March 2010 (UTC)
 * See above. The effective rate per month is 5%/12. Zain Ebrahim (talk) 21:47, 21 March 2010 (UTC)
 * As I wrote above, that was not how my credit card debt used to be calculated every month. But the custom and legislation regarding the advertising and calculation of interest rates is likely to vary from country to country. So best to use APR to make comparisons, although from reading the APR article, even that does not seem too reliable in the USA. 84.13.47.185 (talk) 16:10, 22 March 2010 (UTC)
 * Then your bank didn't use a nominal interest rate - the questioner said "5% interest p.a, compounded monthly", which is a nominal interest rate. Nominal rates are as defined above. Zain Ebrahim (talk) 07:37, 23 March 2010 (UTC)
 * Wikipedia has an article on this. Check out Nominal_interest_rate. Zain Ebrahim (talk) 09:00, 23 March 2010 (UTC)
 * That article has tangled together two things - a) the difference between the stated interest rate and the interest rate after adjustment for inflation (which should be done by dividing not subtracting as the article inaccurately says) and b) a particular way of calculating interest rates. It still seems to be an approximation, perhaps an old-fashioned quirk of the particular textbook cited, and that it is just an approximation is further supported by what the linked article Real interest rate says in the first line, six words in. It was far easier to approximate in the days before scientific calculators rather than taking nth roots, and the practice described looks to be a survival of that. Try looking at Principles Of Corporate Finance by Brealey and Myers for a more modern view. I'm guessing, but perhaps that particular way of calculating interest rates works in the banks favour. 78.149.133.100 (talk) 22:12, 23 March 2010 (UTC)
 * Unfortunately, "nominal interest rates" refer to two entirely different things - that is why I linked directly to that particular section. We're not talking about real rates here. With multiple compounding periods, the definition I described above and the one in the article are NOT an approximation. It's the definition of nominal rates. Zain Ebrahim (talk) 10:22, 24 March 2010 (UTC)
 * I do not believe that the phrase "nominal interest rate" has any second meaning, including the second meaning that you assert. But then banks can do calculations any way they like and give them any names they like. And from what little I've learnt about US banking, it seems that US banks do calculations in what would be considered antiquated ways in the UK, probably because it makes it more difficult for a consumer to unmderstand what they are really getting. 78.146.216.129 (talk) 16:45, 24 March 2010 (UTC)
 * Please stop this. Nominal interest rate DOES have 2 meanings - it's there in the article - regardless of what you might believe. Your incessant refusal to accept this fact does not help our questioner. When you say i p.a. compounded m times per annum, you are referring to a nominal rate in which the effective rate per compounding period is i/m - this is the definition of nominal rates (when the base time unit is one year). Zain Ebrahim (talk) 20:07, 24 March 2010 (UTC)
 * I think your mistake is in not understanding that there are many different ways of calculating interest rates, nearly all of which you could call the nominal rate (by which I assume you mean the apparant rate or whatever else you want to call it). Your error is in trying to dictate that there is one way and one way only of calculating a nominal or apparant rate, which is not true. (Maybe you misread part of a textbook - perhaps what the author meant as a general term, you took as a specific term for the example given). If I may digress slightly, I've noticed with North Americans that you firmly believe that everywhere in the Western world is just like America except for a different language or a funny accent, and you just cannot get your heads around the idea that, even in countries that speak English, the meanings and use of particular words can be different. 84.13.22.69 (talk) 13:13, 25 March 2010 (UTC)
 * Okay, let me try the following approach. Please provide a reference which shows that the "nominal rate" can be treated in the (incorrect) way you describe - and remember, we're not talking about nominal rates in the context of real/nominal rates, we're talking about it in the context of effective/nominal rates. My reference (Wikipedia) disagrees with you and I could find more for you if you want. I fully understand that there are many ways to represent an interest rate - but there is only one way to treat these nominal rates. Btw, I've never been to North America. I live and studied in South Africa and the financial maths I learned was based on the syllabus set forth by the Institute of Actuaries which is in the UK! Zain Ebrahim (talk) 13:57, 25 March 2010 (UTC)
 * Try these http://www.google.co.uk/#hl=en&q=nominal+rate+-wikipedia&meta=&aq=f&aqi=&aql=&oq=&gs_rfai=&fp=23e9d7872b349109 84.13.22.69 (talk) 14:21, 25 March 2010 (UTC)
 * You clearly didn't bother to actually read those results of that google search. The first result deals with nominal rates in the context of inflation - which I specifically stated we're not talking about here and the second result agrees with me! I didn't bother with the rest - you've proved my point for me. Thanks! Zain Ebrahim (talk) 14:30, 25 March 2010 (UTC)
 * The sites do not provide any support for your mistaken idea at all. You are like someone who has learn that the word animal means zebra, rather than being an example of an animal. Now you are saying that the first animal is not an animal (ie not a zebra), that the second site is an animal (although as it includes continuous compounding, that is doubtful), and you refuse to look at any other of the other sites/animals. It is clear that you know you are mistaken but are too proud to admit it. 84.13.34.56 (talk) 13:04, 26 March 2010 (UTC)
 * Nope - you're wrong again. The following is taken from the second site:
 * The formula can be written as:
 * i = n * ( [ (1+r)^(1/n) ] - 1 ),
 * where r is the effective rate, i is the stated rate and n is the number of compounding periods.
 * This is exactly what I've been saying all along. The stated rate, i, is the nominal rate and you can clearly see that i/n is the effective rate per compounding period. That site also deals with continuous compounding which is basically what happens when you let n tend to infinity. Why don't you find a site that actually contradicts me? Zain Ebrahim (talk) 13:13, 26 March 2010 (UTC)

Could someone learn all the notation from all the branches of mathematics in a year?
Say someone was talented at math, but didn't pursue it after high school. They often find mathematical notation on Wikipedia and in some math papers that are of interest to them, but even were they would understand the concepts, they don't know them and certainly don't know the notation. If, fed up with ver seeing alien looking notation that is "Greek to them" they decided to spend a year learning enough math to understand all the notations in every branch of math, could they do it? Or is math now so expansive, with so many branches and sub-branches, that such an endeavor would be akin to not wanting to see a word they didn't kno when reading any major language - which is obviously a totally quixotic undertaking, for which, if it can be completed at all, one year is clearly not sufficient? Thank you. 84.153.237.214 (talk)


 * No. To understand the notation, you need to understand at least some of the maths it represents, and you could never learn all of maths, even to a very low level, in a year (it would be a challenge in a lifetime). There is also the problem of new notation being invented - every time a mathematician comes up with some new maths they invent new notation in order to write it down. I expect more notation is invented in a year than you could familiarise yourself with in a year. There is also the issue of how widespread a notation needs to be before you learn it - do you want to learn notation that is only used in one paper (which is probably most notation). --Tango (talk) 19:58, 20 March 2010 (UTC)

no, I (the op) meant standard notation, notation that many, many (tens of thousands of) mathematicians can understand. The reason I say tens of thousands and not the millions of people who understand more basic mathematical notation, is that I am leaving room for a degree of specialization. However, I am intrigued by your assertion that someone would have to understand the maths they represent to understand the notation. This seems so very false for me. Can you give examples? For example, I studied some college calculus, and I can understand integral and differential notation. However, I failed the course, and I can't take the integral of or derive anything at all. So, by my standard, I understand the symbols used in that course, which probably took me less than two hours to achieve. The vast, vast majority of the course was learning METHODS -- HOW to take derivates and how to integrate. Including many formulas and laws one had to memorize, and so forth. The actual notation itself was very minor. Actually, this example goes lower, we can take it down to algebra and arithmetic. It's super-easy to understand what the log notation is, and then the next 90% of the high school algebra class on logs is about various laws and methods and how to use them. Or a very basic example: the factorial. Every mathematician knows what it means, which takes 10 seconds to learn, and gets used in high school for permutations, and then not so much during the college classes I took. Maybe other math USES it, but there is never anything more to learn about the symbol ! when it means "factorial". Same goes for elementary school when we learn how to turn a(b+c) into a simple sum (no parentheses). It's a method, and 99% of math seems to be about methods, etchniques, proofs, etc. Very little is notation. This is my impression but I welcome your copious counterexamples if I am wrong. Thank you. 92.229.14.140 (talk) 20:22, 20 March 2010 (UTC)
 * When you learn a topic in math there are typically new structures that are introduced, and there are properties of those structures and relationships between those structures that you learn. Knowing the notation for a particular structure is pretty tied up with knowing what the structure is.  For example if you know what $$\lim_{x\to 0}f(x)$$ means, then you probably know what a limit is, and if you know what a limit is then you've doubtlessly learned what the notation $$\lim_{x\to 0}f(x)$$ means.  I guess theoretically you could go about learning all the structures that are studied in various branches of math without learning any properties, but (a) you'd still have to learn a lot of math in the process and (b) I don't know why anyone would want to do that since the structures aren't very meaningful or interesting by themselves.  Typically the structures are motivated by the properties that people want.
 * Just learning all the names is even more useless. What good is it to know that $$\otimes$$ is used to mean "tensor product" if you don't know what a tensor product is?
 * On the other hand I think you could make a point that there is a basic math canon that gets used everywhere like set notation and some other things like that. That isn't too hard to get a handle of, as any course that introduces proof-based math will have to get that stuff out of the way pretty quick. Rckrone (talk) 21:55, 20 March 2010 (UTC)


 * OP, you're right about how it's very easy to know what notations mean, while not being able to use them at all. You might say you can understand adjoints in categories (search adjoint functor), but can't work or even comprehend a trivial question that's asked. Math isn't about learning notation and look cool when you write them; when you actually fully understand them you'll feel the "cool" notations are just like letters a,b,c. When you are more experienced you'll know it normally takes 3 days to understand 2 lines of a really abstract/strange definition. You need think and have lots of mental images in your head to be comfortable with it. So, if you just want to understand symbols, I think you can do it for a particular subject in a month. But you won't even understand the most trivial remarks about it. Money is tight (talk) 06:49, 21 March 2010 (UTC)


 * If you're trying to become a mathematical typist (using TeX) then there are some cheat sheets with the names of all the symbols, and finding the right ones isn't that hard with a little bit of practice. It's more work to learn the weirdness of the software, and writing macros requires some programming-like skill.  But friends of mine have done it without really knowing much math, and it can pay pretty well once you build up a client list.  I can't think of any other reason to be interested in mathematical notation but not in the actual mathematics. 66.127.52.47 (talk) 09:10, 21 March 2010 (UTC)


 * If you want to read things and understand what they mean, it would not be possible to learn the meaning of all the commonly used mathematical symbols and terminology in a year. On the other hand, that's a bit like saying it's not possible to watch every movie ever made in a year. If you did devote an entire year to watching and analyzing a well-selected list of movies, you would come out with both a much deeper appreciation for moviemaking, and a head bursting with new experiences and ideas. If you devoted an entire year to learning a few well-selected areas of mathematics, you would end up knowing vastly more than you do now. What things have you been trying to read that you were not able to? Black Carrot (talk) 23:33, 21 March 2010 (UTC)

Restriction
Given a relation R:
 * 1) Is there a shorter way to express the idea that A ◁ R = R ▷ B?
 * 2) Is there a shorter way to express the idea that both A ◁ R is an bijection injection and A ◁ R = R ▷ B?
 * 3) Is there a shorter way to express the idea that A ◁ R is a function onto B?

HOOTmag (talk) 19:16, 20 March 2010 (UTC)


 * What does the triangle (◁) stand for? --pm a 20:59, 20 March 2010 (UTC)


 * Yeah, I didn't understand the notation either. It seems to be Z notation. So A ◁ R means the relation resulting from the restriction of the domain of R to A, and R ▷ B means the relation resulting from the restriction of the codomain of R to B. —Bkell (talk) 21:05, 20 March 2010 (UTC)


 * Oh, I guess these symbols are described in the Restriction (mathematics) article. Whaddya know. —Bkell (talk) 21:07, 20 March 2010 (UTC)


 * (ec) Are you considering the domain and codomain to be part of the relation, or are you considering the relation to be just the graph? (See Binary relation). If the domain and codomain are part of the relation, then in general A ◁ R = R ▷ B will be false, because the domain and codomain of A ◁ R will be different from those of R ▷ B. If the domain and codomain are not part of the relation, then what does it mean to say that A ◁ R is a bijection? Surjectivity makes no sense unless a codomain is specified. So do you just mean that A ◁ R is injective? —Bkell (talk) 21:04, 20 March 2010 (UTC)
 * Yes, the domain and codomain are not part of the relation. See again the new version of my question above. HOOTmag (talk) 21:21, 20 March 2010 (UTC)