mm-4329 === Subject: Re: Another Bad Induction: All People are of the same gender Bytes: 1981 It is indeed, but it is also a synonym for sex. Check your dictionary. BTW, a similar inductive proof is often used to show that all horses are the same color. See http://en.wikipedia.org/wiki/Horse_paradox. === Subject: Re: Solution Manuals Available Now in PDF! I am interested in obtaining the solutions manual for Engineering Mechanics static's 5th edition Bedford, and the solutions manual for University Physics with Modern Physics 11th edition :author young, the solutions manual and or instructors solution manuals would be greatly appreciated. === Subject: Re: Solution Manuals Available Now in PDF! Two points: (i) The op said Contact bwennny@hotmail.com so why don't you do so? (ii) Learn to snip. Even if there was any point in you posting (which I doubt) there was certainly no need for you to include all of the post you were replying to. -- Remove antispam and .invalid for e-mail address. He that giveth to the poor lendeth to the Lord, and shall be repaid, said Mrs Fairchild, hastily slipping a shilling into the poor woman's hand. === Subject: Re: I need solutions manuals!! CHEATER! you are just going to sell them on the internet. Indian students cheat all the time to get a degree and are worthless on the job. === Subject: Re: I CHALLENGE YOU ALL! P'raps you might consider your own advice? -- The Internet is famously powered by the twin engines of bitterness and contempt. -- Nathan Rabin, /The Onion/ Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Radian equation I need help with two sums. I have the answers but I don't know how to get there. They are supposed to be easy so please help me. 1. tan(-6.b9)+cos(9.b9/4) = 1/A2 2. sin(-3.b9/4)-tan(-.b9/4) = 1-(1/A2) === Subject: Re: Radian equation It's best not to use special characters in ASCII newsgroups. Many people use newsreaders that can't display them. tan(-6 pi) + cos(9 pi/4) = 1/sqrt(2) Hints: The period of the tangent function is pi, so tan(x) = tan(x + k pi), k any integer tan(0) = 0, so what is tan(0 - 6 pi)? The sine and cosine functions have period 2 pi, so sin(x) = sin(x + 2 k pi), cos(x) = cos(x + 2 k pi), k any integer Consider cos(9 pi/4 - 2 pi) sin(-3 pi/4) - tan(-pi/4) = 1 - 1/sqrt(2) Sine and tangent are odd functions, cosine is an even function. Can you work out -sin(3 pi/4) + tan(pi/4) = You shouldn't need it here, but there is the identity sin(t) = cos(pi/2 o t) === Subject: Re: an elementary problem of repeated tangent circles Bytes: 1851 I'm unclear why you would want a recurrence relation, as you can just write down the i'th radius directly. I'll mention inversion. -- Lau AS! d-(!) a++ c++++ p++ t+ f-- e++ h+ r--(+) n++(*) i++ P- m++ === Subject: Re: an elementary problem of repeated tangent circles forehead area center is near x!) and exie and the previous and the previous ---- Galathaea, you've managed to wear out my poor octogenarian brain working out your requested recurrence relationship for general a, b, and a+b radii circles, but here it is. Let a be the smallest radius, b the next, and c = a+b the largest. Let r be the radius currently obtained for an added circle. Then the next radius will be given by the hideous formula: s = b*(a+b)*r*(a*r+b*(a+b)-2*sqrt(b*r*(a+b)*(a-r)))/... ((a+2*b)^2*r^2-2*a*b*(a+b)*r+b^2*(a+b)^2); That defines the recurrence. The zero-th radius, r_0, will of course be r = a. If x-y coordinates are set up with the center of the large circle as origin and x-axis along the line between centers of the three circles, then the center of the r-radius circle is located at: ( (a+2*b)/a*r-(a+b) , 2/a*sqrt(b*(a+b)*r*(a-r) ) This was used to check the correctness of the recurrence formula (not to mention, to obtain it in the first place.) I leave the discovery of sum(r_i) to others with more stamina and courage than I have. Roger Stafford === Subject: Re: an elementary problem of repeated tangent circles I cannot get the sum either. Here are some formulas that might help others. Given four mutually tangent circles R, S, T, U with the circle R is outside S,T,U, then (-1/r + 1/s + 1/t + 1/u)^2 = 2.(1/(rr) + 1/(ss) + 1/(tt) + 1/(uu)). (*) For (r, s, t, u) = (3, 2, 1, u) (-2 + 3 + 6 + 6/u )^2 = 2.(4 + 9 + 36 + 36/(uu)). (7 + 6/u)^2 = 2(49 + 36/(uu)). = (7 + 6/u)^2 + (7 - 6/u)^2 so u = 6/7. If 1/r = 0, and k, l, m = 1/s, 1/t, 1/u then we have a neat formula sqrt(m) = sqrt(k) + sqrt(l). (**) The formula (*) can be verified in this case. Formual (**) can be seen by drawing the three mutaully tangent circles all tangent to the same line, and drawing three right triangles where the hypotenuses are the line segments between the centers, and legs parallel and perpendicular to the line. For (s+r)^2 - (s-r)^2 = 4sr (r+u)^2 - (r-u)^2 = 4ru (u+s)^2 - (u-s)^2 = 4us and therefore 2.sqrt(ru) + 2.sqrt(us) = 2.sqrt(rs). I can prove (*) by inverting in a circle centered at a tangency point into two parallel lines and and two circles tangent to the lines and each other, but it is a bit messy. Suppose the inversion circle has radius a and center O. Measure distance from O. Let C have radius r and center at distance d. Then the radius and distance of the inverted circle are respectively. 1 a.a.r 1 a.a.d - --------, and - -------- 2 dd - rr 2 dd - rr -- Michael Press === Subject: Re: an elementary problem of repeated tangent circles Bytes: 6390 By induction, it's easily proved that r_n=6/(n^2+6) for all nonnegative integers n. Hence, sum(r_n), n from 0 to infinity, equals (1/2)*(1+sqrt(6)*Pi*coth(Pi*sqrt(6))) which is approximately 4.347651084. quasi === Subject: Re: an elementary problem of repeated tangent circles Bytes: 6840 More generally, if the outer circle has radius a+b and the initial inner circles have radii a,b, where a<=b, then r_n = (a*b*(a+b)) / (a^2*n^2+a*b+b^2) for all nonnegative integers n. quasi === Subject: Re: an elementary problem of repeated tangent circles r = a. ----- I notice that the recurrence formula I gave earlier can be simplified to: s = b*(a+b)*r/(a*r+b*(a+b)+2*sqrt(b*r*(a+b)*(a-r))) It still looks rather frightening to me. Roger Stafford === Subject: Re: an elementary problem of repeated tangent circles Bytes: 6501 !! !! I've made a start. Part 3 seems to be the easiest, although I get a !! different answer. !! !! !! Look at any circle that kisses the larger two in this way. Let its !! centre be distance x from the centre of the circle with radius 3, and y !! from the one with radius 2. Then the radius of this circle must be 3-x !! and also y-2, so we know x+y = 5. All of the sequence of circles have !! centres on the ellipse that has poles at the centres of the two original !! circles and passes through their touching point. !! !! The distance between the centres of two successive circles is r_i + !! r_(i+1). The sum of these distances is !! !! sigma(i=0 to oo) (r_i + r_(i+1)) = 2S - r_0 = 2S - 1 !! !! This distance is bounded below by half the circumference of the circle !! with radius 2, above by half the perimeter of the ellipse and further !! above by half the circumference of the circle with radius 3. !! Hence 2 pi < 2S-1 < 3 pi and !! !! pi + 1/2 < S < (3 pi + 1)/2. !! !! The perimeter of the ellipse gives a much closer bound, but I can't !! remember a formula for that and I must leave it for now and get back to !! work. darndarndarndarndarnDARN!!! yes i meant to define C = 2S - 1 and ask about 2 pi < C < 3 pi if i can't even ask the question right why will anyone even want to look?!? anyways that is the geometric argument i wanted which comes in useful when looking at the abstraction to S(a, b) and limits for the rest of the problems there is a trick for 1 that makes the answer come out without too much work but of course the challenge is to find that trick in case anyone is worried that it's just some huge messy expression here is a little hint that may be surprising: r_1 is rational -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: an elementary problem of repeated tangent circles In fact, r_1=6/7. quasi === Subject: Something new for teaching and learning trigonometry Hello. This is my first post here. I am not a teacher by profession, but I have developed something that I am hoping teachers everywhere will find useful for introducing Trigonometry. It is called the TrigRuler and you can get more information about it at - http://www.trigworks.com === Subject: Re: Something new for teaching and learning trigonometry That is really cool! Brian === Subject: Root finding with a vector function for use with a backward Euler integrator. Bytes: 1682 Hey all, I am implementing a root-finding method, essentially for a backward-Euler numerical integrator. So, the general case would be something like I have a function y2 = g( y2 ) where y2 is a vector and g is some arbitrary vector function. I want to solve for y2. I have seen plenty of examples of solvers for cases where y1 and y2 are scalars, but have no idea how to extend that to vectors. I certainly don't think that a bisection method will work in the general case, because the function g might produce roots at different locations for each component of y2. However, in the specific case of the backward Euler method, I might be okay because g( y2 ) = y1 + h * f( t+h, y2 ) where y1 and y2 are vectors, h and t are scalars, and f is the ODE function such that. y' = f( t, y ) So, could someone clarify this? How are these things implemented in practice? Dave === Subject: Re: Root finding with a vector function for use with a backward Euler integrator. Hi Dave, in case you have a system of differential equations, the standard root-finding method is Newton's method. With your notation above, one gets y2_(n+1) = y2_n - (dg/dy(y2_n)-I)^(-1)*(g(y2_n)-y2_n) As a starting guess, y2 from the last time step is taken. For small step sizes h, also the fixed point iteration y2_(n+1) = g(y2_n) will converge. This is easier to implement than Newton's method and needs less storage, but in general reduces the possible step size drastically. Best wishes Torsten. === Subject: Re: Root finding with a vector function for use with a backward Euler integrator. Something wrong with this method. If I have a simple line function g( x ) = m x + b whose root I want to find. So, I set it up like this y_(n+1) = y_n + g( y_n ) However, it seems to me that this won't work if m is large. The large slope will take the value further away from the target. So, am I missing something here? Dave === Subject: Re: Root finding with a vector function for use with a backward Euler integrator. Bytes: 2119 Sorry, I changed my notation a bit. In the above, g is a function whose root I want to find, i.e., I want the x such that g_1( x ) = 0 In previous posts, I was talking about g( x ) being the function such that x = g_2( x ) Well, I can convert from one to the other by doing g_2( x ) = g_1( x ) + x and then solve using fixed-point iteration. At least, that is what I was getting at in the above post. === Subject: Re: Root finding with a vector function for use with a backward Euler integrator. So how do I compute dg/dy(y2_n) in general? I know I can approximate it numerically, but that sounds expensive. Also, what do people generally use to get the inverse matrix? Is the system sparse enough for Gauss-Seidel type stuff? Hmm, how small? What's the restriction on step size? === Subject: Re: Root finding with a vector function for use with a backward Euler integrator. If the right-hand side of your system differential equation is not too complicated, you should provide the partial derivates analytically. Otherwise they are approximated by finite-differences. The inverse matrix is not explicitly calculated. Instead, one solves in each Newton iteration the linear system (dg/dy-I)|y2_n * (delta_y)_(n+1) = - (g(y2_n)-y2_n)) and sets y2_(n+1) = y2_n + (delta_y)_(n+1) For small system, this should be done with direct methods. If the size of the system grows, iterative methods come into play, but this is not an easy task for general linear systems. For the fixed point iteration to converge, a sufficient condition is h*|||df/dy|||_2 <= a < 1. If the spectral radius of the Jacobian of f is big, that's a drastic restriction on the possible stepsize. === Subject: Re: An allegory; Was: All positive integers are equal. On Wed, 08 Aug 2007 08:36:30 -0400, Stephen J. Herschkorn Giggle. Just keep on fighting, then. Have fun. Heh. I see that Han agrees with you. Look up some of _his_ posts and the replies he gets. Show me a post where he's been trying to learn something and he got insulted. (As opposed to posts where he's been explaining that mathematicians have this or that all wrong...) ************************ === Subject: Re: All positive integers are equal. Bytes: 2696 Then you missed the post where Arturo said: Most mathematicians I know would understand the greater of a,b to mean: maj(a,b) = { -- === Subject: Re: All positive integers are equal. Bytes: 3412 No, I didn't miss that. Yes, he said that. He did not (Um, just to make sure you don't misunderstand what I'm saying here, he also did not say Most mathematicians What he said is simply _true_, whether you approve or not: Most mathematicians _would_ understand the greater of a, b to mean exactly what he said. If you'd like to learn to understand mathematical writing you should say oh, I didn't realize that at this point. If you'd rather insist that most mathematcians are wrong in using the language this way, that's your right, but it's extremely silly. ************************ === Subject: Re: All positive integers are equal. days. My association with the Department is that of an alumnus. No, you missed the fact that there is a difference between the relation greater than, and the expression the greater of a,b. === Subject: Re: All positive integers are equal. Bytes: 4187 You're nitpicking over slight and irrelevent differences in verbiage. I again urge you to argue with Eric Weisstein. I have noticed you have made no response to that reference I gave you. Maybe in your world, where you have many, many years of experience you may deem greater in a certain light but you cannot escape the fact that for _most_ people greater than or the greater of or larger than or the larger of implies we are comparing values that are not equal----one value is the greater or larger of the two. This is the standard meaning for many people. You may have a different standard. I see no point further arguing. We stipulate aboyt the OP's original problem depending on the definition of maj(a,b), so there really is not much more to debate except which one's definition is better. I do not intend to debate that, for you would say better could mean either (i.e. they are both just as good.) I think you need to review what er means when appended to words like that. If you feel as strongly as you appear to about your definitions, do a wiki on it or advise Weisstein to change is definition of greater. Or write a text, if you haven't already, using your definitions. -- === Subject: Re: All positive integers are equal. days. My association with the Department is that of an alumnus. No, I am not. There is an important difference between the binary relation greater than, and the expression the greater of a and b. If you think the different is irrelevant, then you should stay away from mathematics. -- Arturo Magidin magidin-at-member-ams-org === Subject: Re: All positive integers are equal. Bytes: 2217 Darrell, for the sake of clarity here is the usage which is almost universal among mathematicians: (2) the greater of a and b means: This is completely standard. Eric Weisstein would certainly not disagree. Best wishes, Peter S. === Subject: Re: The more big list of solutions manual (for all the siubjects) Mechanics of aircraft structures by C T Sun can be purchased from Amazon. If you have difficulty just quote the ISBN: 0471178772. -- Remove antispam and .invalid for e-mail address. He that giveth to the poor lendeth to the Lord, and shall be repaid, said Mrs Fairchild, hastily slipping a shilling into the poor woman's hand. === Subject: Re: The more big list of solutions manual (for all the siubjects) Bytes: 97261 hi, i am searching for Process Dynamics and Control (2nd Ed., Seborg & Edgard) solution manual. thank you for the effort, i appreciate eng doit28@hotmail.com === Subject: Height of a paddle above water There are a couple of things I don't quite get about the following: A paddle on the wheel on a river boat is 7 feet, and the distance from the center of the paddle wheel to the water is 5 feet. Assuming the paddle wheel rotates at 5rpm, and the paddle is at it's highest point at t = 0, write an equation for the height of a paddle relative to the water at time t. So, I came up with 7 cos pi/6 + 5 which is wrong. The reason I use the cosine function is there was another similar problem that used it, but I would really like to understand why it is used. Also, I don't know why I got the pi/5 wrong if the paddle wheel rotates at 5rpm then that's 5*2pi/60sec or pi/6 radians per second. I guess I need to know the key points I missed in the reading material that leads to a solution to the problem. -- OI think Skull and Bones has had slightly more success than the mafia in the sense that the leaders of the five families are all doing 100 years in jail, and the leaders of the Skull and Bones families are doing four and eight years in the White House.O - Ron Rosenbaum === Subject: Re: Height of a paddle above water One problem is that you are missing the variable t completely! Was that a typo? Most of what you have is correct. Apparently you are comparing this to some given answer. Does it make it clear that t is in seconds rather than minutes? That's not mentioned in the answer. In fact, since it gives 5 rpm I would be more inclined to use t in minutes: with t in minutes, the height is given by 7cos(10pi t) + 5. With t in seconds, that would be 7cos((pi/6)t)+ 5. I'm inclined to think your mistake is just forgetting the t! === Subject: Re: Height of a paddle above water G.E. I forgot the t (typo), and since I have posted, I realize the answer was given with the time in minutes, so it was 10pi(t). My only question remaining is, how do you know to use the cosine function as opposed to the sine? -- OI think Skull and Bones has had slightly more success than the mafia in the sense that the leaders of the five families are all doing 100 years in jail, and the leaders of the Skull and Bones families are doing four and eight years in the White House.O - Ron Rosenbaum === Subject: Re: Height of a paddle above water One problem is that you are missing the variable t completely! Was that a typo? Most of what you have is correct. Apparently you are comparing this to some given answer. Does it make it clear that t is in seconds rather than minutes? That's not mentioned in the answer. In fact, since it gives 5 rpm I would be more inclined to use t in minutes: with t in minutes, the height is given by 7cos(10pi t) + 5. With t in seconds, that would be 7cos((pi/6)t)+ 5. I'm inclined to think your mistake is just forgetting the t! === Subject: Can I get the solution manuals for these books? Bytes: 1259 1. An Introduction to Thermal Physics, by D.V. Schroeder (Addison- Wesley, 2000) 2. Mathematical Methods for Physicists(Arfken & Weber, 6th Ed.) If you have, I can afford reasonable prices for them. please email me. === Subject: solution to incropera could u please send me the solutions to INTRODUCTION TO HEAT TRANSFER (THIRD EDITION), BY Frank Incropera and Dawid Dewitt. Its urgent. ragards === Subject: Re: square root Poor Driveby you can't prove anything! Next one please... === Subject: solutions to two books? Do you have solutions to: Analytical Mechanics for Relativity and Quantum Mechanics by Oliver Davis Johns and to: Mathematics for Physicists by Susan Lea Brooks It would be greatly appreciated!!!!!!!!