Wikipedia:Reference desk/Archives/Mathematics/2007 November 1

= November 1 =

test study help?
I'm trying to work this problem to study for a test tomorrow, and I need a tiny smidgen of help.

I'm integrating an expression that eventually reduces to something akin to this (with integration symbols removed): Sqrt((x^2)(b^2 - a^2) + a^4)

The problem specifically says that a > b. So is it OK for me to make the substitution: u = x * Sqrt(b^2 - a^2) du = 1/(b^2 - a^2) dx

...to leave me with: Sqrt(u^2 + a^4)

...even though u would be non-real? Afterwards I can use trigonometric substitution and finish the problem, which reduces to a very elegant and completely wrong answer. Since u is squared immediately it never becomes an overt obstacle (I don't have any Sqrt(b^2 - a^2)'s in my answer), but I think it's messing things up somehow anyway- by looking at the answer it's obvious that the author made a different substitution here, because there's no way he could have used trig sub to come up with the answer he did, and I can use my calculator to confirm that the line directly above the substitution evaluates to the exact same value that the actual answer does (if I pick arbitrary a's and b's- it's a definite integral).

If it helps any, here are the problem and the answers the author and I got: The ellipse (x^2 / a^2) + (y^2 / b^2) = 1, a > b is rotated about the x-axis to form a surface called an ellipsoid. Find the surface area of this ellipsoid. My setup- y = b*Sqrt(1 - (x^2 / a^2)) dy/dx = (-bx) / ( (a^2)Sqrt(1- (x^2)/(a^2)) ) Integral from -a to a of: 2Pi * y * Sqrt(1 + (dy/dx)^2) dx Integral from -a to a of: 2Pi*b * Sqrt(1 - (x^2 / a^2)) * Sqrt( 1 + ((b^2 * x^2) / ((a^4)(1 - x^2 / a^2))) ) dx My answer- (2*Pi*b^2) / a The author's answer- 2*Pi(a^2 + [ (a^2)*B*arcsin( sqrt(a^2-b^2)/a ) / sqrt(a^2-b^2) ])

I think that the author is right and I'm wrong but I'm terrible with Mathematica and the best I can do is Integrate[2 Pi*b*Sqrt[1 - (x^2/a^2)]*Sqrt[1 + (b^2 * x^2)/((a^4) (1 - x^2/a^2))], {x, -a, a}, Assumptions -> (a > b)] Which gives me the horrendous 2 b \[Pi] If[(a > 0 && (a + b > 0 || b >= 0)) || (a >= 0 && b < 0 && a + b < 0), ( Sqrt[-a^2 b^2 + b^4] + a^2 ArcSinh[Sqrt[-1 + b^2/a^2]])/Sqrt[-a^2 + b^2], Integrate[Sqrt[1 - x^2/a^2 + (b^2 x^2)/a^4], {x, -a, a},    Assumptions -> (b < 0 && b < a < 0) || (a > 0 && b == -a) || (b < 0 && b < a < 0)]]

Thanks, -- ⁪ffroth 00:27, 1 November 2007 (UTC)


 * The first thing I notice is that your substitution is wrong, regardless of the values of a and b. If
 * $$u = x \sqrt{b^2 - a^2}$$
 * then
 * $$du = \sqrt{b^2 - a^2} dx$$
 * not $$1/(b^2 - a^2) dx$$ as you say. —Keenan Pepper 02:50, 1 November 2007 (UTC)
 * Oh right duh, I have that right on paper, I don't know why I typed it in like that -- ⁪ffroth 17:09, 1 November 2007 (UTC)

(edit conflict) No u is real since x is squared.

But I'm going to start at the top lets assume you want to integrate the upper curve for the ellipse. so we need to rewrite $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ in y= form

$$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}$$

$$y^2=b^2-\frac{b^2x^2}{a^2}$$

Take the positive root.

$$y=\sqrt{b^2-\frac{b^2x^2}{a^2}}$$

Now you want to do a solid of revolution. with that ellispe. (Creates a specific form of the ellipsoid.) and we want surface area, so it's a Surface of Revolution.

So we also need $$y'$$. Using the chain rule.

$$y'=\frac{dy}{dx}\bigg [ \sqrt{b^2-\frac{b^2x^2}{a^2}} \bigg ]$$

$$y'=\frac{1}{2\sqrt{b^2-\frac{b^2x^2}{a^2}}}\cdot\frac{dy}{dx} \bigg [ b^2-\frac{b^2x^2}{a^2} \bigg ]$$

$$y'=\frac{1}{2\sqrt{b^2-\frac{b^2x^2}{a^2}}}\cdot\frac{-2b^2x}{a^2}$$

$$y'=\frac{-2b^2x}{2a^2\sqrt{b^2-\frac{b^2x^2}{a^2}}}$$

Now for the Integral.

$$S=2\pi\int_{-a}^{a} \sqrt{b^2-\frac{b^2x^2}{a^2}} \sqrt{1+\bigg (\frac{-2b^2x}{2a^2\sqrt{b^2-\frac{b^2x^2}{a^2}}} \bigg )^2} dx$$

That should be your integral. I know, it's ugly but it should simplify pretty nicely. I think it would be best not to simplify anything up to this point (otherwise your liable to miss some patterns).

$$S=2\pi\int_{-a}^{a} \sqrt{b^2-\frac{b^2x^2}{a^2}}\sqrt{1+\bigg (\frac{4b^4x^2}{4a^4(b^2-\frac{b^2x^2}{a^2})} \bigg )} dx$$

$$S=2\pi\int_{-a}^{a} \sqrt{b^2-\frac{b^2x^2}{a^2}}\sqrt{1+\bigg (\frac{b^4x^2}{a^2b^2-b^2x^2} \bigg )} dx$$

$$S=2\pi\int_{-a}^{a} \sqrt{b^2-\frac{b^2x^2}{a^2}}\sqrt{1+\bigg (\frac{b^2x^2}{a^2-x^2} \bigg )} dx$$

$$S=2\pi\int_{-a}^{a} \sqrt{\frac{a^2b^2}{a^2}-\frac{b^2x^2}{a^2}}\sqrt{\frac{a^2-x^2}{a^2-x^2}+\frac{b^2x^2}{a^2-x^2}} dx$$

$$S=2\pi\int_{-a}^{a} \frac{b}{a}\sqrt{a^2-x^2}\sqrt{\frac{a^2-x^2+b^2x^2}{a^2-x^2}} dx$$

$$S=2\pi\frac{b}{a}\int_{-a}^{a} \sqrt{a^2+(b^2-1)x^2} dx$$

From here use a tan and sec trig substitution, be SURE you calculate your new integration limits OR if your not sure how to do that, BE SURE to convert back to x before invoking the integration limits. Hope that helps A math-wiki 04:06, 1 November 2007 (UTC)

Because of the presense of the arctan instead of the arcsin, your answer will appear different, don't be alarmed it doesn't mean that you've made any mistakes. A math-wiki 04:12, 1 November 2007 (UTC)

If you want to use arcsin pull $$-1$$ out of $$b^2-1$$ so instead of $$+(b^2-1)$$ you get $$-(1-b^2)$$. I would actually recommend this on closer examination since that way you can check your answer against your text. A math-wiki 04:18, 1 November 2007 (UTC)


 * I still can't figure out what I did wrong :/ It's probably a matter of being really unlucky and picking a really difficult way of doing it that messes things up in some arcane way.. shortly after the trig sub I run into the integral of sec(theta)^3, which leads me down a page-long rabbit trail O_O -- ⁪ffroth 17:41, 1 November 2007 (UTC)

For integrating powers of secant, try rewriting the integral in two different forms. I only vaguely remember this technique. I would recommend using the arcsin route instead since your text gives an answer with arcsin in it. A math-wiki 21:11, 1 November 2007 (UTC)

equation for series 1,1,3,7,...
I know the next number is (or is greater than) 17, has anyone see this series before?-Dacium 06:24, 1 November 2007 (UTC)


 * Try entering it in http://www.research.att.com/~njas/sequences/. The top hit (http://www.research.att.com/~njas/sequences/A001333) is 1, 1, 3, 7, 17, 41, 99, 239, ... and is given by the numerators of continued fraction convergents to $$\sqrt{2}$$. kfgauss 06:55, 1 November 2007 (UTC)


 * I should add that your sequence could be something quite different, though, so go to the website I gave and enter the start of your sequence and look through a few of the hits. kfgauss 07:01, 1 November 2007 (UTC)


 * I find it interesting that this series could be expressed by the recurrence relation: f(1) = 1; f(2) = 1; f(x) = 2*f(x-1) + f(x-2). - Rainwarrior 09:16, 1 November 2007 (UTC)


 * This recurrence relation arises because $$1+\sqrt{2}$$ is a root of the equation $$x=2+x^{-1}$$, and so the continued fraction representation of $$\sqrt{2}$$ is [1;2,2,2,2,...]. Gandalf61 13:03, 1 November 2007 (UTC)


 * See Pell number for more on this and related integer sequences. Though I agree with some other commenters that there's too little information to be sure which sequence is intended. —David Eppstein 00:25, 2 November 2007 (UTC)

Hate to be a party-pooper, but there's also 1, 1, 3, 7, 19, 51, 139, 379, 1035..., i.e., f(x) = 2*(f(x-1) + f(x-2)) - 1. Though I haven't tested it, it looks like that may be the numerators of continued fraction convergents to e. So there are at least two possible equations that start with those four terms. Grutness...wha?  09:40, 1 November 2007 (UTC)


 * The newton series 1+n&middot;(n&minus;1) reproduces the sequence 1 1 3 7 for n=0 1 2 3
 * 1+n&middot;(n&minus;1)&middot;(1+(n&minus;2)&middot;(n&minus;3)/6) reproduces 1 1 3 7 17 for n=0 1 2 3 4
 * Bo Jacoby 11:11, 1 November 2007 (UTC).

Derivative of an exponential Function
How would I go about finding the derivative of y = 0.8315e0.1958x that equation? Thanks Deltacom1515 13:12, 1 November 2007 (UTC)


 * Perhaps Exponential function might be of some help? —Ilmari Karonen (talk) 13:55, 1 November 2007 (UTC)
 * The Chain rule will also be helpful. Donald Hosek 16:51, 1 November 2007 (UTC)
 * The chain rule is for multiple instances of 'x'
 * View the function as: y=Cc^cx or y=c^x Artoftransformation 10:19, 3 November 2007 (UTC)

math> equations
Hello. I'm having trouble with my algebra II homework. I don't want you to do my entire homework for me, but I do need help on this specific part. The questions are in this format. I seriously have no idea how to do it. Thank you very much! Camillex3 14:58, 1 November 2007 (UTC)
 * If the scientist mixes 50 L of the first solution with 50 L of the other, what percentage does she get? Try to write this down in an equation. If she uses 10 L of the first solution, how much of the other solution does she need, and what is the percentage of the final solution? Now you can substitute 10 for a variable like x, specify that the answer needs to be 35% and you can solve the equation. Feel free to ask if you need more help. risk 15:43, 1 November 2007 (UTC)


 * Or you can set up a pair of simultaneous linear equations and then solve them. For example, if you add x litres of the first solution to y litres of the second solution then you want 100 litres altogether, so


 * $$x+y=100$$


 * and you want to have 35 litres of peroxide per 100 litres of the final solution, so


 * $$0.2x+0.7y=35$$


 * Use you favourite method of solving simultaneous linear equations to find x and y and you are done. Gandalf61 16:43, 1 November 2007 (UTC)
 * The approach that Gandalf has suggested is pretty much what the curriculum is aiming at here. The original poster should have just done a few sessions on solving systems of linear equations in two variables. I would also write that second equation on the first pass as:
 * $$0.2x+0.7y=0.35(100)$$
 * since that makes it clearer where the 35 comes from.
 * It's helpful to write down what all the key values here are as well, so you should write that x is the number of liters of 20% solution, y is the number of liters of 70% solution, 100 is the total number of liters to end with. Then you can see that the first equation is the total number of liters that are being mixed. In the second equation, you have one expression (0.2x) which is the number of liters of peroxide in the 20% solution, another, 0.7y which is the number of liters of peroxide in the 70% solution and then one last, 0.35(100), which is the nummber of liters that you end up with. Donald Hosek 16:49, 1 November 2007 (UTC)
 * I remember these type of problems from 7th grade.. my teacher made us use these diabolical tables that made no intuitive sense whatsoever (sort of like mechanically learning long division) and I never understood it :( I feel for you! Algebra 1 was my most frustrating class until Calculus 2 :P -- ⁪ffroth 17:44, 1 November 2007 (UTC)


 * Fine suggestions; here's another.
 * First, we convert phrases like "20 percent peroxide" into suitable mathematics. If we have 100 L of such a solution (in water), it will be 20 L of peroxide and 80 L of water. Similarly, a 100 L water solution which is 70 percent peroxide will contain 70 L of peroxide and 30 L of water.
 * Now would be a fine time to review George Pólya's advice in How to Solve It. Our article is incomplete; this link provides important details.
 * Since we are asked to make 100 L of a 35 percent peroxide solution, we must mix some amount (call it A) of the dilute (20%) solution with some amount (call it B) of the concentrated (70%) solution to produce a 100 L solution of which 35 L is peroxide. Thus in mathematical terms the goal is to find A and B. We can simplify our task immediately, because we know that A and B together must add up to 100 L; thus knowing A will suffice, because B must then be 100 L−A.
 * Is finding a suitable mix plausible? Yes, because one solution is more dilute than we want, and the other is more concentrated.
 * At this point we could use trial and error.
 * {| class="wikitable" style="text-align:center"

! colspan="3" | Quantities in litres ! A || B || peroxide
 * 100 || 0 || 20
 * 0 || 100 || 70
 * 50 || 50 || 45
 * 75 || 25 || 32.5
 * }
 * In this table we can calculate the 50/50 solution peroxide as half of 20 (10 L) plus half of 70 (35 L). Since the 75/25 solution is too dilute, we can next try half 50/50 (22.5 L) and half 75/25 (16.25 L), producing a 62.5/37.5 mix with 38.75 L peroxide.
 * The good news is, it seems that with patience we can approximate the desired solution. The bad news is, it promises to be slow and never to give the exact answer. But the other good news is that our efforts have not been wasted; if we make a mistake in our attempt at an exact calculation we can check it against our approximations and catch it.
 * To proceed further we should make a small but important change in our setup. Replace the quantity A with the fraction of the total mix it represents, &alpha; = A/(100 L). Now the 75/25 mix corresponds to &alpha; = 0.75. The benefit of this is that we can now calculate the peroxide of any mix directly, because the peroxide of A will be &alpha;×(20 L) and the peroxide of B will be (1−&alpha;)×(70 L). For example, that 62.5/37.5 solution should have 0.625×(20 L) + 0.375×(70 L) peroxide, which is indeed 38.75 L.
 * We are now on the brink of an exact solution. We have a formula in &alpha; that we can equate to the desired peroxide, 35 L. If we are able to solve the resulting equation exactly, we can skip the approximations. Of course, we must remember to convert the &alpha; solution back to the quantities A and B.
 * For someone who has seen numerous problems like this before, a solution method and answer leap off the page. For someone who has not, the experience is far different.
 * The goals of the homework exercise are two-fold. One goal is to train you to be able to recognize and solve such problems quickly; they happen to be the most common problems in all of applied mathematics. A second goal is to learn the skills of problem solving when the solution method is not obvious and when a solution may be difficult (or impossible!) to find. For those whose lives are not pure repetition, such problems — most not involving mathematics — will be pivotal in jobs, and in relationships, and elsewhere. Be grateful to have the opportunity to practice on homework; for when your job or your very survival are on the line in a crisis, the disciplined mental skills you develop now will be invaluable. --KSmrqT 19:31, 1 November 2007 (UTC)
 * In this table we can calculate the 50/50 solution peroxide as half of 20 (10 L) plus half of 70 (35 L). Since the 75/25 solution is too dilute, we can next try half 50/50 (22.5 L) and half 75/25 (16.25 L), producing a 62.5/37.5 mix with 38.75 L peroxide.
 * The good news is, it seems that with patience we can approximate the desired solution. The bad news is, it promises to be slow and never to give the exact answer. But the other good news is that our efforts have not been wasted; if we make a mistake in our attempt at an exact calculation we can check it against our approximations and catch it.
 * To proceed further we should make a small but important change in our setup. Replace the quantity A with the fraction of the total mix it represents, &alpha; = A/(100 L). Now the 75/25 mix corresponds to &alpha; = 0.75. The benefit of this is that we can now calculate the peroxide of any mix directly, because the peroxide of A will be &alpha;×(20 L) and the peroxide of B will be (1−&alpha;)×(70 L). For example, that 62.5/37.5 solution should have 0.625×(20 L) + 0.375×(70 L) peroxide, which is indeed 38.75 L.
 * We are now on the brink of an exact solution. We have a formula in &alpha; that we can equate to the desired peroxide, 35 L. If we are able to solve the resulting equation exactly, we can skip the approximations. Of course, we must remember to convert the &alpha; solution back to the quantities A and B.
 * For someone who has seen numerous problems like this before, a solution method and answer leap off the page. For someone who has not, the experience is far different.
 * The goals of the homework exercise are two-fold. One goal is to train you to be able to recognize and solve such problems quickly; they happen to be the most common problems in all of applied mathematics. A second goal is to learn the skills of problem solving when the solution method is not obvious and when a solution may be difficult (or impossible!) to find. For those whose lives are not pure repetition, such problems — most not involving mathematics — will be pivotal in jobs, and in relationships, and elsewhere. Be grateful to have the opportunity to practice on homework; for when your job or your very survival are on the line in a crisis, the disciplined mental skills you develop now will be invaluable. --KSmrqT 19:31, 1 November 2007 (UTC)

Thomas & Finney's
can anyone please tell me where can i find the solution maunual online in pdf format( i knot it exists in electronic form) of book called CALCULUS by thomas & finney.. i would be very grateful if some one locates it.. —Preceding unsigned comment added by 203.128.4.242 (talk) 16:58, 1 November 2007 (UTC)


 * As far as I know, "Calculus and Analytic Geometry : Student Solution Manual, Part 1 (Paperback)

by George B. Thomas (Author)", ( odd numbered problems) and the Teachers edigtion( all problems) are not avaible in pdf form for the 9th edition. I checked online, and at a bookstore. Artoftransformation 10:28, 3 November 2007 (UTC)

Quickest possible way to find GCF and LCM
Hi. No, this is not homework. I'm just asking so I could do it quicker in the future. Now, I know a few ways to do this already. One way is to list all the factors or as many multiples as needed, and find the greatest/lowest common one. This can be time consuming and involve too many steps. Another way is to list a factor tree for the numbers until there are only prime numbers left, and circle the common ones (eg. if there two numbers to factor, and there are two 3's on one number and five 3's on the other, we circle two 3's from each). We then multiply all the circled ones (only take them from one of them, eg. if there is 3, 2, and 2 circled on both numbers, you do 3, 2, and 2 once, not twice), to find the GCF, and multiply that by all the uncircled numbers to find the LCM. This, however, can be time-consuming if the numbers have a large amount of prime factors, and can be very easy to mess up. So, I'm asking for the quickest possible way to figure this out. This must be: So, is there a way? Also, while we're at it, are there any prime numbers exsisting if you include its negative integer factors? What about 1? Thanks. ~ A H  1 (TCU) 17:44, 1 November 2007 (UTC)
 * able to be done without a calculator
 * possible to do for an average person/question using only thinking (mental math) in under one minute
 * be done without using any trial-and-error
 * give a correct answer for all sets of positive numbers when we want to find the GCF and LCM
 * be easy to remember and not easy to mess up
 * will not take up too much room if it is done on pen(cil) and paper
 * be easy for even elementary students to understand
 * will not take too much time even when there is a large number of multiples/factors that may need to be stated
 * There's always the Euclidean algorithm which involves successively finding remainders. It's reasonably quick for medium sized numbers and very fast for very large numbers. For example to find the GCF of 56 and 80, we find the remainder when we divide 56 into 80 and get a remainder of 24. Then we take 24 into 56 and get a remainder of 8, then 8 into 24 and find it divides evenly so the GCF is 8 (I usually don't use it for numbers under 100 myself as I've got those factorizations memorized). Donald Hosek 18:17, 1 November 2007 (UTC)
 * Once you know that, how to find the GCF, I am pretty sure you can find the LCM of two numbers X and Y as X * Y / GCF(X, Y). For more than two I think you can do this recursively. Baccyak4H (Yak!) 18:31, 1 November 2007 (UTC)

One small mistake there Baccyak4H, it's GCF(X,Y)^2 since both X and Y have GCF(X,Y) as a factor, their product has two of that factor. So for 56,80 Your answers are 8 and 70, I find that Euclidean Algorithm to be very efficient, I will probably start using it regularly. A math-wiki 21:23, 1 November 2007 (UTC)
 * Not quite. Let d be the GCF of x and y, then $$x=ad$$ and $$y=bd$$ for some integers a and b with GCF of 1. The LCM of x and y is $$abd=\frac{(ad)(bd)}{d}=\frac{xy}{d}$$. Donald Hosek 21:53, 1 November 2007 (UTC)
 * Seriously, LCM means "least common multiple", which among other things, means it is a multiple of both numbers. 70 is a multiple of neither 56 nor 80. -- Meni Rosenfeld (talk) 05:48, 2 November 2007 (UTC)
 * Umm... I think you may have missed a few of my points. It must be:
 * possible to do for an average person/question using only thinking (mental math) in under one minute,
 * be easy for even elementary students to understand.
 * First of all, I find parts of this very helpful, especially the $$\frac{xy}{d}$$ part because I understand it right away, and it works. I find the Euclidean algorithm part helpful too, but since I find the article overly confusing, I need more examples. So, the way I percive this is, you take the two numbers, then you divide the smaller number into the bigger number, then you take the remainder, divide that into the smaller number, take the remainder, and divide that into the previous remainder? What if that doensn't divide evenly? What if it divides evenly prior to the place when it divided evenly in the example? Do these two examples, $$\frac{xy}{d}$$ and Euclidean algorithm, work if there are more than two numbers to find the GCF and LCM of? I found that in many previous ways I've tried, they have often failed for more than two numbers. I will probably need more examples of how it works when there are more than two numbers. Also, what does a comma do in an equation? I have a calculator that does the comma thing, but I can't find it right now. Also, what about the prime number part of the question? Usually, when I'm given a formula, I can see how and why it works, not just simply, "it works". That's how I found Pathagorean theorm (sort of) using plain logic and some testing. However, I can't visualize how these work at the moment, which is why I ask. So, could someone help simplicify or give more examples of these formulas? Can these be done efficiently without pen(cil) and paper? Thanks. ~ A H  1 (TCU) 22:46, 1 November 2007 (UTC)


 * (edit conflict)
 * The Euclidean algorithm is thousands of years old; how hard can it be? Given two integers a and b, with a the larger and with b nonzero, suppose integer d divides both, so that a = dm and b = dn for some integers m and n. Then d also divides the difference a−b, since it is d(m−n). Therefore we lose no possible divisors if we replace a by the smaller number a−b. If we do this repeatedly, then eventually a will become smaller than b, but non-negative. If it becomes zero, then b is a common divisor, and in fact the largest since we discarded no potential factors. Otherwise, swap a and b (so that a is again the larger and b nonzero) and continue.
 * To speed up the calculation, suppose that we may subtract b from a for a total of q times before the difference r is less than b. Then we may write a = qb+r, where q is the integer quotient of a/b and r is the remainder. With this in mind, we take out the repeated subtraction and replace it by finding the remainder under integer division.
 * The extended algorithm finds not only the GCD, d = (a,b), but also two integers i and j such that ai+bj = d. To this end, the extended algorithm makes use of the quotients q. --KSmrqT 23:27, 1 November 2007 (UTC)


 * The simplest form of the Euclidean algorithm uses subtraction. Division is a shortcut - it speeds it up and gets the right answer, but it's a bit less obvious why. Here's subtraction. AC - BC = (A-B)C, so the difference of two numbers is divisible by any factor they share. On the other hand, AC + BC = (A+B)C, so the sum of two numbers is also divisible by any factor they share. Let's say we want the greatest common factor of M and N. Call there difference P, so M - N = P. P is divisible by their greatest common factor, but on the other hand, M = N + P, so M is divisible by any common factor of N and P. In other words, the greatest common factor of M and N is the same as the greatest common factor of N and P. This can be written gcf(M,N) = gcf(N,P). The comma just means I'm listing two numbers. It's also the case that gcf(M,N) = gcf(M,P), so gcf(M,N) = gcf(N,M-kN), where k is any whole number. Taking the remainder after division is the same as subtracting one number over and over again from the other and writing down what's left, so the same rules apply. Black Carrot 23:02, 1 November 2007 (UTC)


 * Ok, so I sort of see how that works. However, is (N,M-kN) the same thing as (N,(M-(k*N))) or as (N,((M-k)*N)), where I put as many unnesecary parentheses and stuff as nessecary? Also, can someone provide me with more examples of the division way, and do the three ways/parts ($$\frac{xy}{d}$$, division way of Euclidean algorithm, and the subtraction way) work when there are more than two numbers where we want to find the GCF and LCM? Also, just curious, although this will likely not have a practical usage, do these formulas work if the numbers from which we want to find the GFC/LCM of were negative integers; what about if one of them were? What about the prime numbers question; if prime numbers became composite from the negative factors, would they be in the same category as normally composite numbers, or would those be bumped up even more? Thanks. ~ A H  1 (TCU) 23:26, 1 November 2007 (UTC)
 * You can't aribtrarily put parentheses anywhere. $$(N,M-kN) = (N,(M-(k*N)))\;\!$$ because these are the same thing according to the order of operations, but in general $$M-kN \neq (M-k)N$$ and $$(N,M-kN) \neq (N,(M-k)*N)$$. You do have, however, $$(N,M) = (N,M-kN)\;\!$$ because of the properties of the GCD (which I think is a more common and correct term than GCF) which were mentioned above.
 * You don't need to worry about negatives - that is, the GCD is the same if you take the negatives of some numbers. For example, (-24, 10) = (24, 10) = 2. Prime numbers don't become composite from negative factors (7 has divisors 7, 1, -1, -7 and it is doing just fine as a prime).
 * It looks like you have understood the algorithm correctly, so you can come up with your own example for practice. Euclidean algorithm also has at least one example. I'll give another one to get you started: (126, 30) = (30, 6) = (6, 0) = 6.
 * About your eariler comment... You do realize that you have asked for a method which satisfies quite a bit of requirements, right? You are lucky for the existence of one which does satisfy all of them. In particular, it can be done mentally under a minute, for 3-digit numbers, provided you know how to divide; it is easy to learn how to do, and easy to understand if taught properly. -- Meni Rosenfeld (talk) 05:48, 2 November 2007 (UTC)
 * I like that phrase: "as many unnesecary parentheses and stuff as nessecary". --Lambiam 11:36, 2 November 2007 (UTC)

The fastest technique I've come across for finding the gcd is this old Chinese one. Pick two numbers, say 132 and 84, and you want to find the gcd. Write them one in a column:

132

84

Subtract the smaller from the larger, and replace the larger with the difference:

132->48

84--->84

Repeat, until the number on top and on bottom are the same:

132->48-->48-->12-->12-->12

84--->84-->36-->36-->24-->12

When you get the same number on top as on bottom, that's your gcd.

SmaleDuffin 20:36, 2 November 2007 (UTC)


 * Hello?! Perhaps you should read what's been written before. This is precisely (the non-division version of) the algorithm we have all been discussing. I even gave a proof of correctness! --KSmrqT 21:39, 2 November 2007 (UTC)


 * Hi. I've found your answers very helpful, but there's stll something I can't figure out. How do you find the GCF or LCM for three or more numbers? I can't seem to find a straightforward way to find this using Euclidean algorithm, so is there a quick way to find this for many numbers? I know that, if the numbers aren't divisible by each other at all, the GCF is 1 and the LCM is all the numbers multiplied together. However, is there a quick was to find it for 3 or more numbers? Also, I find it almost impossible to do questions like "find the GCM for 29482940298 and 284724874" in my head, and very time-consuming on paper. Thanks. ~ A H  1 (TCU) 14:19, 4 November 2007 (UTC)
 * There may be better ways to find the GCD of multiple numbers, but the best I can think of, which was mentioned briefly, is to do it "one at a time", taking advantage of the fact the the GCD is "associative", in the sense that $$(a,b,c) = ((a,b),c)$$ and so on. For example, suppose you want to find the GCD of 210, 330, 462 and 770. You first calculate (using the Euclidean algorithm) (770,462) = 154, then (154,330) = 22 and finally (22,210) = 2, which is the final result. You can do the same for the LCM, but I think you'll have to do the computation from scratch - there is no apparent way to use the GCD of all numbers to easily calculate their LCM.
 * An average human cannot even subtract these two numbers mentally, did you expect a method that will allow you to find their GCD? That's what electronic computing devices are for. On paper, a skilled arithmetician can do them in less than 10 minutes, which is much faster than the average case for any other technique (for these particular numbers it will be relatively easy to use their prime factors, assuming you have a table of prime numbers). -- Meni Rosenfeld (talk) 14:54, 4 November 2007 (UTC)

51-gon??
I've recently had some fun using a geometry program I downloaded to construction Regular Polygons. I have constructed all through and including a 51-gon, but before I tried my method to construct the 51-gon, which I verified works I did some searching on the internet to see if there were any documented methods of construction, ... without result. I'm curious if there are documented work of construction of 51-gons, to compair with my method. Thanks A math-wiki 21:28, 1 November 2007 (UTC)
 * See Constructible polygon. Carl Friedrich Gauss proved that a regular 51-gon can be constructed with Compass and straightedge (51 = 3&times;17). PrimeHunter 22:00, 1 November 2007 (UTC)
 * Altough Gauss proved it could be constructed, he did not provide a method.
 * (after edit conflict) Have you read the article "Constructible polygon"? This provides a recipe for constructing a regular 51-gon (actually a constructable pq-gon where p and q are distinct Fermat primes) which entails beginning with a circle in which you will inscribe a regular 17-gon and an equilateral triangle with a shared vertex and from there, selecting the two vertices which are 360/51 degrees apart and using that angle to place the remaining 32 points on the circle for your 51-gon. The heptadecagon article gives the construction for the 17-gon. Donald Hosek 22:07, 1 November 2007 (UTC)

That was the generalization I found and used for the 51-gon, but how about constructing a 257-gon. How would one actually construct it. I know it's complicated but I am insufficiently versed in using $$\Phi (k)$$ and other more advanced techniques to figure it out on my own. Thanks A math-wiki 01:52, 2 November 2007 (UTC)
 * Two places to start would be with the ratio 360/257, and hunting down a book on Advanced Algebra, which points to the root equations that are used as a proof of non-contruction. Artoftransformation 10:34, 3 November 2007 (UTC)
 * The angle $$\frac{2\pi}{257}$$ is important, but of course, the ratio 360/257 itself is meaningless, as degrees are an arbitrary unit. -- Meni Rosenfeld (talk) 00:03, 4 November 2007 (UTC)