Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

September 9

Acceleration dependent on position

I've been trying to solve this problem: Given a particle with mass m, initial position (0,0,0), and initial velocity ${\dot {x}}_{0}{\bar {i}}+{\dot {y}}_{0}{\bar {j}}+{\dot {z}}_{0}{\bar {k}}$, find equations for the particle's position and velocity if the force acting on it is $(4-y){\bar {j}}$. I've figured out that since there is no x or z acceleration, the velocity in those directions will be constant, so all I need to find is velocity in the y direction, ${\dot {y}}$. From F=ma I know the acceleration in y: ${\ddot {y}}=(4-y)/m$ My first thought was just to integrate both sides with respect to t, but I can't figure out how to integrate the y term on the right side. Any suggestions? — jwillbur 16:24, 9 September 2008 (UTC)

See ordinary differential equation. --Trovatore (talk) 16:45, 9 September 2008 (UTC)

Bah, I knew I was missing something obvious. Thank you, Trovatore. — jwillbur 19:35, 9 September 2008 (UTC)

About statistics

Q. What is the target population? What property a representative sample is expected to hold? —Preceding unsigned comment added by 89.108.14.159 (talk) 17:51, 9 September 2008 (UTC)

This sounds a lot like a couple of questions out of a homework assignment to me, but in the interest of good faith I will provide some hints and motivating examples. Firstly, a "target population" is something that assumes that you are interested in some piece of information. If you know what information you want to know, then you should know what your target population is. Suppose you are going to run a survey to find out about the study habits of students at a university. Who is the target population? As for what property a representative sample should hold, consider an unrepresentative sample - in this case, let's say I just ask everyone in my statistics class. Why is this sample not representative of my target population? Confusing Manifestation(Say hi!) 00:55, 10 September 2008 (UTC)

Index numbers

What are index number and their uses in business? —Preceding unsigned comment added by 89.108.14.159 (talk) 17:53, 9 September 2008 (UTC)

This doesn't really sound like a maths question. An index number is a way of referring to an item or concept or ... when you have several of them and want a short way of consistently referring to each thing (or group of things). A car registration number is an "index" supplied by the government to uniquely identify a particular vehicle (eg XY08ABC may refer to "Ford Ka chassis number 123456789 Engine number 987654321". It can be used for almost anything.

Alternatively an index number may be a statistic. Eg the cost of a typical week's shopping. Usually these are set to 100 when they start, so an index of 110 means that the cost has gone up 10% with respect to the start date. These are good for observing trends and may be of help in future planing. -- SGBailey (talk) 10:02, 10 September 2008 (UTC)

From my undergrad in finance, index numbers are often used in discussions about inflation. i.e. what does an apple cost today in 1945 dollars. You can learn more at Price index and List of price index formulas Sentriclecub (talk) 15:00, 10 September 2008 (UTC)

statistics

Argue that descriptve statistics and inferential statistics are must for each other.

Explain the difference between population and statistical population? —Preceding unsigned comment added by 89.108.14.159 (talk) 17:57, 9 September 2008 (UTC)

Sounds like homework to me... Try searching Wikipedia for those phrases and see if you can't do it yourself. (If that fails, try google). --Tango (talk) 18:17, 9 September 2008 (UTC)

Three homework questions in a row? Come on, you're not learning anything by just asking others, and there's no point in studying a course if you're not learning anything. -- Aeluwas (talk) 11:42, 10 September 2008 (UTC)

September 10

Query regarding divisibilty

We have got some basic ways to find out the divisibility of an integer(irrespective of its size) by 2,3.. til 10 (within 10). But is there any basic ways to get it applied for divisibility by 7. —Preceding unsigned comment added by Bbsxyz (talk • contribs) 14:44, 10 September 2008 (UTC)

There's nothing very easy. The equivalent of the usual test for divisibility for 3 is that if you take the units digit of your number, add 3 times the tens digit, add 2 times the hundreds digit, subtract the thousands digit, subtract three times the tenthousands digit, subtract 2 times the hundredthousands digit, add the millions digit, and so on (repeating the pattern 1, 3, 2, -1, -3, -2) then the result of this sum is divisible by 7 if and only if the original number was. That probably isn't any quicker by hand than just long division by 7 though. Algebraist 14:50, 10 September 2008 (UTC)

See Divisibility rule#Divisibility by 7. Gandalf61 (talk) 14:54, 10 September 2008 (UTC)

That's correct, and fast. A slower method, but easier to remember is to expand the 'number value' taking 3 instead of 10 as the system base.

Say a number 2175 = 2×10³ + 1×10² + 7×10¹ + 5×10⁰

Calculate its 3–base value: 2175 → 2×3³ + 1×3² + 7×3¹ + 5×3⁰ = 2×27 + 1×9 + 7×3 + 5×1 = 89

Iterate the method: 89 → 8×3¹ + 9×3⁰ = 8×3 + 9×1 = 33

33 → 3×3¹ + 3×3⁰ = 3×3 + 3×1 = 12

12 → 1×3¹ + 2×3⁰ = 1×3 + 2×1 = 5

Number 5 is not divisible by 7, so the initial number 2175 is not, either. Also 89, 33 and 12 are not divisible — all of them give the remainder of 5.

The coefficients given by Algebraist and Gandalf above are simply equivalents of powers of 3 with respect to divisibility by 7 (we say 'congruent modulo 7', see Modular arithmetic), chosen as close to zero as possible, so that numbers calculated faster converge to (-6, 6) interval. --CiaPan (talk) 07:19, 11 September 2008 (UTC)

What are the rules of treating differentials?

I learned calculus from a Stewart book which looks down on using dummy variables as actual terms like dv = a dt

I am wondering if someone can point me to a page (here or on the internet) that outlines the rules for how engineers and physicists treat the differentials as meaningful terms. My Stewart book said that the dt that appears after the integrand is only to remind us how Leibniz developed them from the concept of Reimann sums.

I'm specifically looking at

$dv=-v_{rel}{\frac {dM}{M}}$

$\int _{v_{i}}^{v_{f}}dv=-v_{rel}\int _{M_{i}}^{M_{f}}{\frac {dM}{M}}$

I don't understand how in the top equation, I can integrate both sides between different limits of integration? In my newbie understanding, that is like dividing one side of an equation by 12 candelas, but dividing the other side by 13 pascals.

My second question, is changing the limits of integration from v_i and v_f to M_i and M_f. I understand Integration by substitution clearly as it is taught in my Stewart book. I could understand if both sides were integrated between limits of integration if the terms had the same dimensions (velocity and velocity) but then one of the integrals changed the integrand from a velocity to a mass (and in complying with the FTC-2 of the subsitution rule, also changing the limits v_i to M_i and v_f to M_f).

I guess my apprehension is worrying about if the left side and right side will balance.

Third question is 5 second easy one. When you evaluate a definite integral, does the answer always have the same dimension as the integrand multiplied by the dummy variable? This is a pattern I've noticed.

Thanks, I'm a contributor to the Differential article and other pages where I can help out with clearity and links and navigability. Also, I've read 7 articles trying to find my answer, and have tried looking online. Thanks in advance. My only 3 questions at the ref desk have all been about differentials. Sentriclecub (talk) 14:56, 10 September 2008 (UTC)

Here's my work so far...

$\int _{a}^{b}dv=(b-a)\cdot y$

and I also have this worked out

$-v_{rel}\int _{a}^{b}{\frac {dM}{M}}=-v_{rel}\cdot (\ln b-\ln a)=-v_{rel}\cdot \ln {\tfrac {b}{a}}$

So as you can see, I'm stuck on only seeing the dM and the dv as a dummy variable only, and want to use y = f(x) lingo that is in my engrams. —Preceding unsigned comment added by Sentriclecub (talk • contribs) 15:23, 10 September 2008 (UTC)

You can think (extremely unrigorously) of the "dv" or "dM" or whatever as being an infinitesimal change in v or M, or whatever, so dM is probably the mass of an infinitesimal amount of the item in question. You can manipulate them in much the same way as any other variable and it usually works just fine, but just don't let a mathematician see you doing it! (And sanity check your answers just in case you've found one of the few cases where it doesn't work.) As for adding limits on both sides, I think they need to be the same limit, just in terms of the appropriate variable (so your limits are probably the values at t=0 and the values at t=T, or something, you just use the velocity at those times on one side and the mass at those times on the other). And yes, the dimension of d(Something) is the same as the dimension of the something and integration just means adding lots of them together, so that doesn't change the dimension. --Tango (talk) 15:31, 10 September 2008 (UTC)

If you like to be rigorous, just treat dx and dy as new variables. The rules of algebra are the addition rule: d(x+y) = dx+dy, the multiplication rule: d(x·y) = x·dy+dx·y, the power rule: d(x^y) = y·x^y−1·dx+x^y·log(x)·dy, and the constant rule: dk = 0 if k is constant. Refer to differential algebra in case mathematicians object, as they sometimes do because they got scared by the unrigorous interpretation of differentials as infinitesimals. Note that the limits refer to the dummy variable behind the d. Explicitely:

$\int _{a}^{b}dv=\int _{v=a}^{b}dv=\int _{x=a}^{b}dx=b-a$

$\int _{v=v_{i}}^{v_{f}}dv=-v_{rel}\int _{M=M_{i}}^{M_{f}}{\frac {dM}{M}}$

Note that M_i=M(v_i). Bo Jacoby (talk) 16:52, 10 September 2008 (UTC).

Thanks so much for the help. I knew I understood it when you adduced Note that M_i=M(v_i) because I have spent the last 90 minutes, and my brainstorming was that a and b are points that have a lot of information. I conjectured a function that assigns a to v_i and a function that assigns a to M_i and then integrated this function from initial to final.

I reasoned that I am still only integrating from a to b and the only difference is that I'm now integrating between f(a) and f(b) a new function which assigns to each element f(c₁) in a set I to an element c₁ in a set II

I concluded that every value between a and b has a corresponding M_i and a corresponding v_i. Thus some value between a and b has a corresponding M_j and v can be thought of as a function of M since every M_j has only one corresponding v_j. Thanks very much! I bookmarked this great explanation from my userpage, and I tried a problem from my book and it was finally easy, clear, and simple. Sentriclecub (talk) 17:35, 10 September 2008 (UTC)

You've certainly got your work cut out to distinguish clearly between the various forms of differentials and derivatives. Offhand I can think of things like partial derivative, exterior differential, covariant derivative, never mind the boundary operator looking like a partial derivative. Using differentials is good I think but there is some problems especially with twice differentiating. In differential forms d² is zero, and for ordinary differentials y"dx² isn't the same as d(dy) which if one works it out would come to y"dx²+y'd(dx). Dmcq (talk) 18:04, 10 September 2008 (UTC)

Thanks dmcq for the input. I trust the reasoning behind Stewart's exclusion of this information from his formal calculus textbook. I purposely tried to avoid learning the stuff about shorthand differentials treating them as autonomous terms, but I only did so to make better sense of my physics textbook. Plus the public lectures from MIT's opencourseware, is taught by professor Walter Lewin who uses this treatment of differentials in upcoming electromagnetism lectures. Both these were a factor, but I wish I had the time-machine to go back to age 18 and majored in physics and not finance. I'm now halfway to 30 and have a lot to catch up! Sentriclecub (talk) 18:27, 10 September 2008 (UTC)

Ulp I thought I say the quote marks in what I wrote but obviously I didn't and they have turned into the thing to make italic writing. I'll fix that last line and write it here again. For ordinary differentials y"dx² isn't the same as d(dy) which if one works it out would come to y"dx²+y'd(dx). One has to resist the urge to use d²x here. There's also Lie derivatives if you're interested. Dmcq (talk) 18:52, 10 September 2008 (UTC)

Dmcq points at the difficulty in differentiation twice using differential notation. The clue here is that when x is an independent variable, then dx is constant, and so ddx=d²x is zero. So y=x² gives dy=d(x²)=2·x·dx and d²y=d(dy)=d(2·x·dx)=2·d(x·dx) =2·(dx·dx+x·d(dx))=2·(dx²+x·d²x) =2·dx² and d²y/dx²=2. Bo Jacoby (talk) 09:41, 11 September 2008 (UTC).

True if you can keep the business of what is independent clearly in mind. In fat you might like the full expression because simply by setting d² to zero so y is the independent variable you get a nice way of deriving d²x/dy² from the value of d²y/dx² Dmcq (talk) 18:38, 11 September 2008 (UTC)

Stewart is a fanatical adherent of a religion that holds that all calculus textbooks must be identical in all essentials (and you make yours brilliantly different by doing something really original such as putting the Mean Value Theorem in Section 4.2 instead of Section 4.1, which has been done by only 950 previous authors, or slightly changing the colors of the figures, and that although one must not attempt logical rigor, nonetheless the order in which one presents the topics must be what would make sense if one did. That religion forbids being honest about the Leibniz notation, and says one must tell students that they're no good and just historical curiosities. Michael Hardy (talk) 05:53, 14 September 2008 (UTC)

mathematician name for famous subset of whole #'s 1,2,3,5,8,13,21  ?

Trying to help my Daughter with no Text book —Preceding unsigned comment added by 24.215.67.178 (talk) 21:26, 10 September 2008 (UTC)

Fibonacci. -- The Anome (talk) 21:30, 10 September 2008 (UTC)

There is a web-site, the On-Line Encyclopedia of Integer Sequences (link), where you can type a sequence such as yours, and get it identified if it is in any way notable. typing in your numbers, returns

Fibonacci numbers: F(n) = F(n-1) + F(n-2), F(0) = 0, F(1) = 1, F(2) = 1,

Note, also, that the first two numbers of the sequence are missing in your question, it should be 0,1,1,2,3,5,8,13,21,... Except for these two first numbers, every number is the sum of the two preceding ones. See Fibonacci number. --NorwegianBlue ^talk 23:44, 10 September 2008 (UTC)

Actually, even those two numbers are the sum of the two preceding ones. The Fibonacci sequence is easily extended in the negative direction. Fredrik Johansson 12:40, 12 September 2008 (UTC)

September 11

Origin Of The Concept Of Zero?

From my understanding, the concept of a zero has been conceived in many different cultures at different times, but the earliest known reference to the concept "zero" was from the Dogon Tribe of Mali (West Africa).

The term meant "dead" or "from dead", in context it meant "nothing", it was symbolised by a mark made by the thumb. It was later adopted by various middle eastern countries and the thumb print was replaced by the symbol we know today (0), as it was easier to write (it is still representative of a thumb print).

The term "dead on" (meaning on point) also has relations to the Dogon meaning of zero.

For some reason the knowledgeable minds of the world are reluctant to give any credit to sub-Saharan Africa for the intellectual development of mankind.

Intellegent people have lived in Africa from even before the dawn of civilisation (European standard), so it's not such a far stretch to think that most of the worlds inventions (like the 12 volt batteries found in Egypt) originated in Africa and have been copied, modified, or re-invented in many parts of the world.

Mankind is an intelligent creation. Worldwide, unrestricted from time and text.

Now your probably thinking "that's interesting, but where's the question know it all?"

Well here it is (they are);

I know that the Dogon Tribes "zero" pre-dates the other "zero's" but am not sure of the date it was first documented or concieved. This is why I am not updating the page in question.

Does anyone have info on this matter? and if so, why no mention of the "Dogons"

I know the Dogons did not date and time everthing they did, but they deserve a mention.

You will also find no reference to the "Phonicians" when it comes to "Phonetics" (also an African influence on the world) and many other subjects.

I generally find Wikipedia to be a good impartial source of information, but the problem of "crediting Africa" or "giving credit where credit is due" is very common, can this be fixed?

G.Logic —Preceding unsigned comment added by Godlogic (talk • contribs) 13:04, 11 September 2008 (UTC)

The word Phoenicia is unrelated to "phonetics", which comes from the Greek word for "sound" (like telephone, microphone, etc.). So I do not understand what connection between the Phoenicians and phonetics you are referring to.

The issue of a systemic western bias on Wikipedia is a serious issue: see Wikipedia:bias for information. However, I do not believe this particular matter you bring up is an example of that systemic bias. I haven't heard of a Dogon zero, and couldn't find mention of it with Google; from what I recall of the book "Zero: The biography of a dangerous idea", a history of the origin of the number, no mention was made of the Dogon tribes. An abstract concept like the use of zero as a number does not arise in isolation, but comes about from using numbers in a sufficiently abstract way to necessitate its invention; so I doubt the Dogons did, in fact, invent zero.

It would be helpful if you can provide any recollection of where you heard of this connection between the Dogon tribe and the number zero. I understand from what you say that you do not have an encyclopedic reference (and am grateful that you refrained from modifying the article for that reason), but even an uncyclopedic source of information would be of help in tracking down an encyclopedic source. Eric. 65.96.172.100 (talk) 14:58, 11 September 2008 (UTC)

Followup to an earlier question

I have a quick followup to my original question. How would I go about determining the probability that an element is part of "Collection B" if it has the three most significant features, in this case Feature 1, Feature 2, and an absence of Feature 5 (with individual probabilities of 0.96±0.03, 0.88±0.04, and 0.66±0.08 respectively)? Many thanks for your help! But I Played One On TV (talk) 18:52, 11 September 2008 (UTC)

Just treat the absence of a feature as a feature and you've got your original question. And as Bo Jacoby said in answer to the original question you have to start counting up the particular feature set rather than using the aggregated statistics as the various features may not be independent of each other, and the dependence may not show up until you put three or more features together - though that is luckily not very common. Dmcq (talk) 07:18, 12 September 2008 (UTC)

See [[1]] Bo Jacoby (talk) 08:35, 13 September 2008 (UTC).

September 12

how to predict the next number with 4 digits

what is the percentages to predict a totally exact number each time?

-if someone of you know anything details aboutthe prediction or formula applied to find a exact number with 4 digits each time,please let me know .thanks for sharing.

If I understand your question correctly, you are asking what the probability of guessing one correct number in 10000 numbers (0000 to 9999). Assorted conditions apply - the main one being that the next number is independent of the previous number. If that is the case then the answer is 1/10000 as a probability and 100 * 1/10000 as a percentage. -- SGBailey (talk) 11:26, 12 September 2008 (UTC)

What did they use before graphing calculators?

Pre-calc seems to depend on graphing calculators. But obviously they weren't always around. What did they use before the graphic calculator for pre-calc? I'm assuming the abacus and slide rule weren't sufficient. ScienceApe (talk) 21:43, 12 September 2008 (UTC)

What part of precalc requires a graphing calculator? I'm not quite sure what precalc consists of (I was educated in England), but I think you can do it all with pencil and paper. Nothing at school relies on complex calculations. Of course, there were people in pre-calculator days who had to perform complex calculations, beyond the abacus and slide rule. They used pen, paper and a lot of time and effort. Algebraist 21:48, 12 September 2008 (UTC)

I don't know. The professor just said that there are parts of the class that required a calculator, and that such problems were impossible to do without the use of a calculator (even for him). It just made me wonder how they were able to do such calculations before the graphing calc was in common usage for highschool/college students. ScienceApe (talk) 03:32, 13 September 2008 (UTC)

See Human computer for one alternative. A friend of mine asked a class when they thought the industrial revolution was, wide variations, one answer was 1960 :) Dmcq (talk) 21:51, 12 September 2008 (UTC)

True, but I don't think professional human computers were ever made available to students in high schools. Algebraist 21:57, 12 September 2008 (UTC)

Oh okay then, books of tables. Students used thin booklets of four or five figure tables, professionals used books with 7 figure tables and you learn rules to interpolate for the last couple of digits. You could get huge great big thick books with all sorts of tables including even tables of random numbers. Dmcq (talk) 22:08, 12 September 2008 (UTC)

Here's a classic professional mathematicians book of tables Handbook of Mathematical Functions: with Formulas, Graphs, and Mathematical Tables Dmcq (talk) 22:16, 12 September 2008 (UTC)

I have to say that in my time doing maths (junior school to degree level), I only ever used a graphics calculator once or twice, and that was when we were being taught how to use them. They're banned in most exams, and I can't think of any good reason to use one really. Most graphs you don't need to actually know the exact curve, just a sketch and a few quick calculations are enough. -mattbuck (Talk) 23:10, 12 September 2008 (UTC)

That's weird. I've taken pre-calc three times so far (nothing to do with my intelligence, just school politics), and all of them require a graphing calculator. I'm not talking about Math in general, it seems only pre-calc uses it really. ScienceApe (talk) 03:37, 13 September 2008 (UTC)

Could you give us an example of a question from your textbook, or maybe from an exam, that is allegedly impossible to answer without the assistance of a graphing calculator? One of the old-timers will probably be able explain how it would be done with more primitive tools. --tcsetattr (talk / contribs) 06:16, 13 September 2008 (UTC)

Graphing calculators are great for getting a quick rough idea of what a function does and finding out roughly where the axis intercepts are. Once you've got that, it's much easier to know what you need to do work things out more precisely (for example, you know how many intercepts there are so you know when to switch from finding more to proving there are no more). I can't think of anything that absolutely requires a calculator, but it is much quicker that working it out analytically from scratch, of plotting it by hand. --Tango (talk) 08:23, 13 September 2008 (UTC)

Reading this thread, I was also puzzled by which part of pre-calculus actually requires a graphing calculator, as opposed to it being a handy convenience tool. Digging around a bit, I found this question paper from this pre-calculus course outline, which includes the following question:

“ Use your graphing calculator to sketch the graph and label the vertex, y-intercept, and x-intercepts:

1. y = -2(x + 3)² + 7
2. 4x + y – 2x² = 9

”

IMHO, the student who algebraically determines the co-ordinates of the vertex and intercepts and confirms these by sketching the graphs by hand has actually learned something about parabolas, quadratic equations and how graphs change under certain linear transformations of co-ordinates. But the student who follows the instructions, copies a graph from the screen of their graphing calculator, and reads off and writes down the approximate locations of the vertex and intercepts has only learned how to use a graphing calculator and has learned little or no mathematics. It is a perfect example of a problem where the route to the answer is more important than the answer itself. Gandalf61 (talk) 11:41, 13 September 2008 (UTC)

That's a question intended to test your ability to use a graphing calculator. If that's specifically on the sylabus, presuming they expect you to need to know it later. Judging by the name, I expect pre-calc is meant to prepare you for calculus - you don't need a graphing calculator for any calculus course I've done... --Tango (talk) 13:47, 13 September 2008 (UTC)

No no no, pre-calc has absolutely nothing to do with Calculus nor does it prepare you for Calc (don't ask me why they call it pre-calc, I don't know). I'm guessing you don't have pre-calc in UK or where ever you are from, but it's here in America. Pre-calc is basically the study of functions, graphs, and advanced algebra. ScienceApe (talk) 17:38, 13 September 2008 (UTC)

...and how to use graphing calculators, apparently ;) Gandalf61 (talk) 21:08, 13 September 2008 (UTC)

Early pocket calculators were expected to be used for hard stuff like 34201·4837, but actually they are used for computing 3+18, which is no longer done by hand. Teachers expect students to use graphing calculators for sketching y = −2(x + 3)² + 7, but probably it is used more for sketching y = 2 and y = x and the like. Bo Jacoby (talk) 13:36, 13 September 2008 (UTC).

In the past before learning calculus a student would just have to plot each y point for different x values. They would then learn as part of the calculus course how to find the minimum and maximum points and possibly inflexion points which would make sketching the graph and finding the zeros much easier. Dmcq (talk) 14:42, 13 September 2008 (UTC)

I certainly did that during my A-levels, 3 or 4 years ago - maybe it's just the US that teaches using graphics calculators rather than actually plotting graphs. (I did have a graphics calc for my A-levels, but I don't remember using it very often.) --Tango (talk) 14:45, 13 September 2008 (UTC)

Here in Ontario, Canada, we learn to graph lines, parabolas, and high-order functions by hand in grades 9, 10, and 12, respectively. Graphing calculators are sometimes used for advanced algebra because the other steps involved are more important than solving the zeros of an equation like y=2x^5+3x^4-23x^3+9x^2-9x+200.

I'm assuming that pre-calculus corresponds to Ontario's advanced functions course, which is about high-order polynomials, continuity, zeros, asymptotes, and the like. Advanced functions is a pre-requisite for the calculus & vectors course, so in this sense it can be termed "pre-calculus". --Bowlhover (talk) 03:13, 14 September 2008 (UTC)

(American, currently undergraduate.) I can't really speak for the effectiveness of a particular Pre-Calc course, but the purpose of precalculus is to get you ready for calculus by introducing or reviewing things you'll need to know about, like limits, the transcendental functions, and basic use of cartesian coordinates. If it's been decided you'll need to be able to use a calculator in calculus, that's a reasonable tool to teach the use of in precalculus. It can't replace any of the other knowledge you'll need, other than the use of older tools like log tables, but it's not an unreasonable thing to put a little time into, just to make sure everyone's on the same page. Most of the things I've used a calculator for in school involved precise calculations that couldn't be conveniently done by hand, in answer to questions like, "What is the measure of angle A to three significant digits?" or "What is the determinant of this matrix?" Neither of those makes use of an advanced calculator's graphing, approximation, or symbol manipulation capabilities, and no question a national test will ask does either. If you can't do it with a scientific calculator and some thought, it shouldn't come up. I remember teachers using the displays of graphing calculators to teach theory, since it's nice to have a fast and accurate picture of what you're learning about, but that doesn't come up on a test. That still doesn't explain the ubiquity of the TI-83 in our curriculum, which I blame on secret kickbacks from calculator distributors. Black Carrot (talk) 06:16, 14 September 2008 (UTC)

One other thought, on rereading the second Original Post. It's not entirely outside the realm of possibility that your teacher isn't actually good at math. Just something to think about. Black Carrot (talk) 06:19, 14 September 2008 (UTC)

So we seem to have reached the conclusion that the parts of a pre-calculus course for which you absolutely require a graphing calculator (rather than just a numerical scientific calculator) are just those parts that teach you how to use a graphing calculator. Yes ? Gandalf61 (talk) 08:20, 14 September 2008 (UTC)

No, I don't think so. The professor said parts of the exam require the use of a graphic calculator. ScienceApe (talk) 14:17, 14 September 2008 (UTC)

Since you failed to provide an example of a question that can't be answered without the graphing calculator (and the only example given by anyone else was "use your graphing calculator to do something that's already easy enough to do with pencil and paper alone") you forfeit the right to object to the conclusion. --tcsetattr (talk / contribs) 20:17, 14 September 2008 (UTC)

Before graphing calculators they used common sense (except for people who didn't use common sense). Look: If you don't understand graphs of polynomials, rational functions, exponential and logarithmic functions, etc. without a graphing calculator then you don't understand, and a calculator won't change that. Calculators are not supposed to be a substitute for using your head. Nor are they supposed to be an anesthetic. Michael Hardy (talk) 23:13, 14 September 2008 (UTC)

September 13

Calculus Question

Integral of 1 over radical (6x-x^2) —Preceding unsigned comment added by 169.229.75.140 (talk) 01:10, 13 September 2008 (UTC)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn how to solve such problems. Please attempt to solve the problem yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Thank you. Algebraist 01:13, 13 September 2008 (UTC)

I agree with Algebraist that this looks like a homework question, but I'm feeling generous so I'll give you a hint: My first step would be to use substitution to get rid of the 6x term, it should be easy from there. Good luck! --Tango (talk) 08:19, 13 September 2008 (UTC)

In my schooldays, many years ago, the integral of 1/(2ax-x^2)^0.5 was a standard form. Times change.—81.132.236.163 (talk) 14:25, 13 September 2008 (UTC)

It may well be in the back of a standard calculus text, I don't honestly remember - I'm pretty sure it is once you do the substitution, though (that's the easy bit I referred to!). It may depend on the textbook - if I were writing it, I'd keep the list short and expect readers to do the simple conversions to turn their problem into a standard one! --Tango (talk) 14:37, 13 September 2008 (UTC)

I was curious, so I dug out my calculus textbook from the bottom of the box of notes I brought it home for the summer in (Salas, Hille and Etgen if you're interested) and what you get after the substitution is on the back cover, the OP's question isn't. It seems Salas and I are of like mind! --Tango (talk) 14:42, 13 September 2008 (UTC)

Now the OP has had time to think. First substituting x=3+3t and later substituting t=sinv gives

(6x−x²)^−1/2dx = (6(3+3t)−(3+3t)²)^−1/2d(3+3t) = (18+18t−(9+18t+9t²))^−1/23dt = (1−t²)^−1/2dt

= (1−sin²v)^−1/2d(sinv) = (cos²v)^−1/2cosvdv = dv

So

(6x−x²)^−1/2dx = d(arcsin(x/3−1))

Bo Jacoby (talk) 01:24, 14 September 2008 (UTC).

OK, at least one nonsensical answer appears above and at least one that is needlessly complicated. Let's try being straightforward: if you're not instantly thinking completing the square then there's a gap in your understanding that you badly need to fill. So:

${\begin{aligned}6x-x^{2}&=-(x^{2}-6x)\\&=-(x^{2}-6x+9)+9{\text{ (add }}-9{\text{ to complete the square and}}\\&{}\qquad \qquad \qquad \qquad \qquad \qquad {\text{compensate by adding }}9)\\&=-(x-3)^{2}+9\\&=9-(x-3)^{2}\\&=9\left(1-{\frac {(x-3)^{2}}{9}}\right)\\&=9\left(1-\left({\frac {x-3}{3}}\right)^{2}\right)\\&=9(1-u^{2}).\end{aligned}}$

If

$u={\frac {x-3}{3}}$

then

$du={\frac {dx}{3}},\quad 3\,du=dx$

so

$\int {\frac {dx}{\sqrt {6x-x^{2}}}}=\int {\frac {3\,du}{3{\sqrt {1-u^{2}}}}}=\arcsin u+C=\arcsin \left({\frac {x-3}{3}}\right)+C.$

Pretty routine once you've practiced a bit. But if you don't recognize this immediately as a completing-the-square problem, you need to learn this: the purpose of completing the square is always to reduce a quadratic polynomial with a first-degree term to a quadratic polynomial with no first-degree term. Michael Hardy (talk) 03:27, 14 September 2008 (UTC)

What are the chances

My wife went to work the other day and my son joked and said "We may never see her again." We laughed a bit but I got to wondering, what are the chances that we may never see here again? How does one go about establishing the probability that any one particular event like that is canoccur? I think that you would start let's say with the number of highway fatalities in the area in any one week or year, and then perhaps calculate the number or vehicals that travel on the roads that make it home safe every day and so on. I never studied statistics or probability - so I am curious how the though process in this sort of event is calcutate. Just so we are clear - I do wish my wife to come every day.142.68.147.80 (talk) 17:43, 13 September 2008 (UTC)

Yes, you would need to looks at how many fatalities there are doing a particular activity and divide it by how much that activity is done. You could improve the accuracy by restricting it to people like your wife (does she have a good no claims bonus? If so, you should look at how many people with good no claims bonuses die on the roads divided by how much they drive). You should probably base it on miles driven, rather than number of cars, although the risk may well not be proportional to the length of the journey. You would need to gather lots of data to get accurate statistics. Either that, or you could try googling - someone has probably done such a study already for most common activities (well, they certainly will have studied it in order to work out insurance premiums, but they may not have published the results). --Tango (talk) 18:05, 13 September 2008 (UTC)

And then double this number, because you yourself could die too. Sentriclecub (talk) 09:39, 14 September 2008 (UTC)

You can't necessarily double it, since you'll be doing different things and are a different person, so have different risks. Also, you can't just add the probabilities, you have to minus off the chance of your both dying otherwise it will be counted twice (this becomes obvious if the change of each of you dying is more than 50% since adding them together would get a probability of over 100%). --Tango (talk) 12:55, 15 September 2008 (UTC)

September 14

Validity of proof

Is this proof valid? If so, can you help me find a reference for it on the Internet? --Bowlhover (talk) 02:00, 14 September 2008 (UTC)

What's highlighted in green is accurate, and is a straightforward enough application of basic principles that it doesn't need a reference. As a formatting consideration you might try taking the fractions out of their math brackets and just writing them normally, p²/q². Black Carrot (talk) 05:50, 14 September 2008 (UTC)

"Since all numbers which can be represented by p/q are by definition rational, the above implies that all rational square roots are roots of a ratio of two perfect squares," should be rephrased somewhat, the implication is in the wrong direction. Maybe, "Since all rational numbers can by definition be represented in the form p/q, the above implies that all rational square roots are roots of a ratio of two perfect squares." Black Carrot (talk) 05:54, 14 September 2008 (UTC)

The proof does depend on the unstated assumption that gcd(p²,q²)=1 if gcd(p,q)=1. This is true in a unique factorization domain such as Z, but is not true in all integral domains. Other proofs make this dependence clear by using unique factorisation directly. Gandalf61 (talk) 08:36, 14 September 2008 (UTC)

Yep I agree with both contributors above, the proof is flawed. The sentence "p²/q² can be assumed to be in lowest terms" is where it all falls down but it was heading the wrong way from the beginning. Dmcq (talk) 08:50, 14 September 2008 (UTC)

Is it necessary to deal with any numbers except integers? A perfect square and its square root must be integers, and p and q can be assumed to be. --Bowlhover (talk) 14:27, 14 September 2008 (UTC)

If I may interject two points. First of all, I find the section too detailed for inclusion in the main article. Perhaps there is enough material scattered about to start an article on the square root of rational numbers. There are a number of proofs indicated in the square root of 2 article, and I would rather see the section in irrational number kept brief and in summary style. Second, the lengthy proof is not necessary. One can prove considerably more in much less space by appealing to Gauss's lemma (for which there already is an article): Any rational number which is an algebraic integer is a (rational) integer. siℓℓy rabbit (talk) 14:51, 14 September 2008 (UTC)

Would like a math expert to look over my experimentation with a simple physics equation

$f_{k}={\frac {joules}{meter}}={\frac {d}{dx}}E_{th}$

Frictional Force $=f_{k}=Newtons=Joules/meter=Force$ shows that the kinetic frictional force is the derivative wrt x of internal kinetic energy of an isolated system as the block undergoes a displacement, x. Since $f_{k}$ is a constant, and is equal to $\mu _{k}\,$*N (where N = the normal force) then $\mu _{k}\,$ can be thought of as...

Help me from here, please. I want to continue the line of thinking that $f_{k}\,$ is a derivative of internal kinetic energy.

A month ago, I realized that $\mu _{k}\,$ is a percent relative the kinetic frictional force to the Normal Force, and this insight really helped me understand everything in physics much better. i.e. $\mu _{k}\,$=.40 means that of the Normal force, the kinetic frictional force is about 40% the Normal force.

I also really liked the algebraic proof that $\mu _{s}\,$ is the solution to the equation $\mu _{s}=\tan \alpha \,$ where alpha is the angle of incline which a block begins to slide, from rest position.

So today, I'm trying to round out my collection and would like an insightful way to represent an intuitive relationship between $\mu _{k}\,$ to work or internal-kinetic-energy.

So, ignoring angles of incline, and ignoring mass and gravity (let them be 0, m₁, and g all respectively) is there any way for me to algebraically rearrange variables or shuffle equations to get a neat relation between \mu_k and work or internal kinetic energy? I haven't found one, so if one doesn't exist, are there any other outside the box ways to look at these relationships between variables?

When I usually try to creatively play with formulas, if the term is ${\mu _{k}}^{2}$ or something like a difference of terms, such as-- ${\mu _{k}}$ initial minus ${\mu _{k}}$ final, terms like these have more flexibility and you can apply more algebra to get more new equations from old ones.

Thanks in advance for any input. My goal is to learn physics slowly and very analytically, always looking for ways to inject redundant mathematical thinking into my notes that I keep. Sentriclecub (talk) 12:23, 14 September 2008 (UTC)

addendum from the kinetic friction page...

When an object is pushed along a surface, the energy converted to heat is given by:

$E=\mu _{\mathrm {k} }\int F_{\mathrm {n} }(x)dx\,$

where

F_n is the normal force,

μ_k is the coefficient of kinetic friction,

x is the coordinate along which the object transverses.

I may have wrongly presumed that calculus leads to nowhere. (I'm stronger in algebra than calculus) Sentriclecub (talk) 12:29, 14 September 2008 (UTC)

numerical analysis

plz help me solve this example with detailed solution. Find a root of the equation x sin x + cos x = 0, using Newton-Raphson method. Miral b (talk) 12:42, 14 September 2008 (UTC) Retrieved from "http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics" Hidden category: Non-talk pages that are automatically signed

Rearranging the equation gives x = −cot(x). Looking at the graphs of y = x and y = −cot(x) you can see that there are an infinite number of solutions. As you only need to find one solution, you could find the solution that lies between 3π/4 (where x sin(x) + cos(x) = 0.9588...) and π (where x sin(x) + cos(x) = -1) - I am assuming x is in radians here. Starting from either of these initial values, Newton-Raphson converges quickly to a solution, giving at least 8 decimal places in just 4 iterations. I set it up in Excel - alternatively you could use your favourite programming language, or do it with a calculator in less than 10 minutes. Gandalf61 (talk) 13:25, 14 September 2008 (UTC)

The function $f(x)=x\sin x+\cos x$ is an even function: $f(-x)=f(x)$. The Taylor series

$x\sin x+\cos x=1-\sum _{k=1}^{\infty }{\frac {(-x^{2})^{k}}{2(k+1)(2k)!}}=1+{\frac {1}{2}}x^{2}-{\frac {1}{8}}x^{4}+{\frac {1}{144}}x^{6}-{\frac {1}{5760}}x^{8}+\cdots$

contains only even powers of x. So it is simpler to set $t=x^{2}$ and solve the nth degree equation

$1-\sum _{k=1}^{n}{\frac {(-t)^{k}}{2(k+1)(2k)!}}=0$

for some sufficiently big value of n.

The two smallest solutions to x sin x + cos x = 0 are

$x\approx \pm 1.19968i$

The smallest real solutions are

$x\approx \pm 2.79839$

In order to reach this accuracy you need n ≥ 9. The J-code used is %:_1 _2{1{>p.((*1&o.)+2&o.)t.2*i.9 , and the result produced is 0j1.19968 2.79839. Bo Jacoby (talk) 22:14, 14 September 2008 (UTC)

Is impulse related to work?

When an object is pushed along a surface, $\mu _{\mathrm {k} }\,$ is the ratio of the energy converted into heat vs this denominator:

$\mu _{\mathrm {k} }={\frac {E_{th}}{\int F_{\mathrm {n} }(x)dx\,}}$

How do i verbally say the denominator? Is it the indefinite integral of the Normal Force (as a function of x) with respect to x?

I am picturing an example. A constant force is applied to block A which slides 2 meters on flat surface, then up a 60 degree incline for the final 8 meters.

I see a lot of new ways to further explore uses of $\mu _{\mathrm {k} }\,$ now, but needed to know what the denominator is in spoken words.

If this block is pulled by a string in such a way that the block has constant velocity of 1 meters per second (and the string always pulls parallel to the block's momentum), then I could integrate the normal force as a function of time, then convert it into Joules. This would involve solving for the impulse, then converting an impulse into work. I never covered subjects from Calc_II. Is it straightforward to convert an impulse to an amount of work, given this example? I have read the articles on these subjects, but they get too complicated too fast, and I get overwhelmed. I should be able to better attempt my first question if I am not afraid of making a calculus goof. Sentriclecub (talk) 13:28, 14 September 2008 (UTC)

For the denominator, you can just say "the integral of the normal force with respect to x" in most cases. Saying "integrating with respect to x" usually implies that every non-constant term in the integrand is somehow a function of x. You shouldn't really say "indefinite integral" in this context since, strictly speaking, the integral should have lower and upper limits (respectively, the initial and final positions of the object being moved). In the specific example you gave, it is very easy to convert the impulse into work since you know everything about the applied force -- in fact, if you use all SI units, the two quantities are numerically equal since you set the velocity to be 1 m/s. With Calc_1 knowledge, you should be able to quickly verify this using the equations $J=\int Fdt$ and $W=\int Fdx$ where J is the impulse and W is work. The complete answer to your question of whether there is a general relationship between impulse and work is more complicated. You definitely need to know the actual masses of the objects in question. I might explain this more in depth later, but for now I'll give you some helpful information that may lead you to work out the complete answer yourself. First, note that $K={\frac {p^{2}}{2m}}$ where K is the kinetic energy and p is the momentum. Second, review the work-energy theorem, which states that the total work done on an object is equal to the object's change in kinetic energy. 97.90.132.94 (talk) 07:08, 15 September 2008 (UTC)

Thanks, and I hope you'll explain more, after a few days (link now available from my user page ). I have a huge love for mathematics. I spend additional time, when learning physics, to try under understand the equations from a extremely mathematical point of view. I love working through proofs, and I will start working immediately on the part you gave me. That term with momentum^2, I've never seen before, so I'll jump right on it. I'm self taught calculus, so if I'll need more calculus knowledge to understand it, then thats a helluva motivation! Thanks very much! Sentriclecub (talk) 14:56, 15 September 2008 (UTC)

earliest date of use or description

I began using computers to verify results of logical equations I solved previously by hand in about 1963. Then in about 1978 I found a computer method for reducing logical equations to minimum form in “Digital/Logic Electronics Handbook” by William L Hunter (pages 112-113, Tab Books, Blue Ridge summit, PA) ISBN 0830657740 ISBN 0830657746 ISBN 9780830657742 called the Harvard chart Method of logical equation reduction. In 1981, I published a modification of the method to reduce multi-valued equations to minimum form as a computer program here. In doing the modification I may have become unconsciously aware of the ability to count and to sort sets and multisets using an indexed array as demonstrated here.

Nonetheless, I did not become consciously aware of this method until at least 1995, followed by its copyright and online publication in 1996 here and again in 2006 here. Consequently, I am searching for any publication prior to my own which might describe or demonstrate this method to count and to sort sets and multisets or to find the date the Counting sort was first described or published with its original definition found here and the date the Pigeonhole sort was first described or published with it's original definition found here. 71.100.10.11 (talk) 14:04, 14 September 2008 (UTC)

I looked at your BASIC program and am puzzled by the statement 20 a(a)=a(a)+1 . a cannot be an array and an index at the same time. The claim that your sorting routine is the fastest in the world is not proved. Usually the speed of sorting routines is measured by the asymptotical behavior of the time consumption as a function of file size when the file size is big. Bo Jacoby (talk) 11:11, 15 September 2008 (UTC).

I think the original program was written in Zbasic. Yes, I know that speed claim is based on assumption since for one thing I did not create Instant sort, the hardware version, until Nov. 1996. ~ WP:IAR 71.100.4.227 (talk) 16:49, 15 September 2008 (UTC)

I seem to half remember that in some early variants of BASIC (BBC Basic?) ordinary variables, typed variables functions and names all had separate address spaces, so you could do something like that (and have the string A$, integer A% and so on to do something else with). With the source in memory and 16k to play with people would actually use all the single-character variable names they could in a large program. Edit: Yes, I found a reference:

Note that in BBC BASIC the different types of variable are completely independent. For example, the integer variable list%, the numeric variable list, the string variable list$ and the arrays list%(), list() and list$() are all entirely separate.

From [2] -- Q Chris (talk) 12:57, 15 September 2008 (UTC)

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)

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 goes exactly with my point.