mm-1879 === I'm stuck with this integral, int( sqrt(1+sin(x)) )dx Yes, it can be integrated exactly by hand. The substitution z = tan(x/2) gives dx = 2 dz / (1 + z^2), sin(x) = (2z) / (1 + z^2) and will transform the integral to one that can be done with a u substitution and one that can be done with a further trig substitution. The answer is: 2*(sin(x)-1)*(1+sin(x))^(1/2)/cos(x) + C For a discussion of that substitution, see, for example,: http://www-math.mit.edu/~djk/18_01/chapter24/section03.html Or just note sqrt(1+sin(x)) = sqrt(1-sin^2(x))/sqrt(1-sin(x)) = sqrt(cos^2(x))/sqrt(1-sin(x)) = cos(x)/sqrt(1-sin(x)); we've used sqrt(cos^2(x)) = cos(x), which is OK if x is in [-pi/2,pi/2], say. The integration is now very simple: Let u = 1-sin(x), then du = -cos(x)dx, etc. Answer: -2*sqrt(1-sin(x)) + C. This agrees with your answer on this interval (you have a superficial problem with cos(x) in the denominator, but it's not really a problem). This problem is more interesting than I thought. We have an antiderivative on [-pi/2,pi/2] - or any other interval where gives 2*sqrt(1-sin(x)) as an antiderivative. If we define F(x) = -2*sqrt(1-sin(x)) on [-pi/2,pi/2], F(x) = 2*sqrt(1-sin(x)) on [pi/2,3pi/2], then F is an antiderivative for sqrt(1+sin(x)) on [-pi/2,3pi/2]. This F still agrees with your answer on this interval. However, your function can't be an antiderivative for have the same property. Hence no 2Pi-periodic function, in particular the one you found, can be an antiderivative on all of R. To find an antiderivative everywhere, let I = [-pi/2,3pi/2], and define G on the interval I + 2Pi*n by G(x) = 4sqrt(2)*n + F(x - 2Pi*n); here n is any integer. Then G'(x) = sqrt(1+sin(x)) for all x in R. The constant 4sqrt(2) is the integral of sqrt(1+sin(x)) over I; the constants 4sqrt(2)*n insure that the pieces fit together smoothly. This problem is almost too interesting; I wonder what the intention of the author of this exercise was. posting-account=KR2cuw0AAACZ_86pfubjOKsQkAVb6Rpe Forgot all about that! Dave Hello all, when one knows the result of int(f(x).dx) and one knows also y=h(x) is it then possible to directly calculate with I know in normal cases it would be logical to determine first f(y) and then integrate this function, but this is practically impossible (software trouble). Thx for any pointers or advise, Tim it works out fine! May I catch your question this way? f and g two real known functions , g(0)=0 and g(x)=Int( 0 to x f(u),u) ,h(u) being an other real function can we express Int( 0 to x f(h(u)),u) from g(x) ? I think of two cases: 1Á)h(u)=a*u+b ,you've got a solution like g(x)/a.. 2Á)h and f defined such as h(u)=f^[-1](a*f(u)+ m(u)) ,int(mu)=M(u) known, we get Int( 0 to x f(h(u)),u)=Int( 0 to x af(u)+m(u)), = a*g(x) + M(x) - M(0) To be clear about notation let's not use h for two different functions. If y = h(x) then write x = h^(-1)(y) so h^(-1) is the inverse function of h. If I understand correctly, you are wanting to calculate: int( f ( h^(-1)(y) ) dy Make the substitution y = h(x) so dy = h'(x) dx and you get: int( f (h^(-1)(h(x))) h'(x) )dx = int( f(x) h'(x) )dx Does that help you? --Lynn space. Determine Whether R^3 is spanned by the vectors Vector Space R^3 Vectors: | 1| |-1| | 0| | 3| |-1| | 0| |-1| |-2| | 2| | 3| | 5| | 2| space. days. My association with the Department is that of an alumnus. I suspect you mean How, not Who, in the subject... Since every spanning set must contain a basis, one way would be to see whether there is a subset of 3 vectors (3 being the dimension of R^3) which are linearly independent. Alternatively, for each vector check to see if it is a linear combination of the previous ones; if so, discard it and keep going. If not, go on to the next vector. If at any point you are left with fewer vectors than the dimension of the space, then it cannot span. Finally, if you don't know about dimensions yet, notice that you can span R^3 if and only if each of (1,0,0)^T, (0,1,0)^T, and (0,0,1)^T is in the span of the vectors (you'll have to explain why if this is homework). So solve the system by row reducing the matrix that has the vectors you want in the first four columns, and these vectors in the other three. If there are solutions to all three, then your set spans; if not, then it does not. I'll give you a quick hint: the third vector is equal to the first plus the second; so anything which is in the span of all 4 is in the span of the first, second, and fourth. (Be sure you explain why). Do -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu Theory of quadruplet patterns of floral functions Functions : R = 1 + 1*sin(p*A/q)^M and R = 1 - 1*sin(p*A/q)^M R = 1 + 1*cos(p*A/q)^M and R = 1 - 1*cos(p*A/q)^M Diagrams : Graphs of these four functions are congruent without phase difference. Condition : Constant p is odd integer and q = 4*n for n is integers. Variable A = (0, 2*q*pi). Graphic example Web site - www.b192907.com/mathgraph 1. Open application program 2. Select topic 18 quadruplet patterns 3. Give data 1,1,5,4,1 for constant a,b,p,q,M R = a + b*sin(p*A/q)^M, .... 4. It will plot four diagrams of the functions Suggestion and comments are welcome Reference : The quadruplet patterns of the first kind were given in Pattern Mathematics by Dr. Shih (Fig 5.19, p92). The book is free if you can come to pick it up. Dr. Keh-Gong Shih 11 Lantana Terrace, Dartmouth Nova Scotia, Canada. B2X 3L3 posting-account=8jB0Vw0AAACJmu18OPPftv6igzbfKPoN phase integers. Hello all, Perhaps this sequel might be of interest: On Ramanujan's Other Pi Formulas http://www.geocities.com/titus_piezas/ramanujan.html It discusses two more of his pi formulas and we come up with new ones, four derived by this author. It's the fourth file, in html format. --Titus It depends how quickly he stops. A body falling from rest in a constant gravitational field will gain kinetic energy during the fall of: KE = mgh where m is the mass of the body, g is the gravitational acceleration and h is the distance the body falls. Now on the deceleration side, the work done on the body as it stops is W = (Fav) s where Fav is the average force acting on the body and s is the distance it takes for the body to stop. But the body is stopped when its KE reaches zero -- in other words, when the work done decelerating the body equals the KE it had when the body started decelerating, or: KE = (Fav) s mgh = (Fav) s Fav = mg(h/s) In other words, the *average* stopping force on the body is the weight of the body times the ratio of the falling distance to the stopping distance. So if a 250lb person falls 5ft and then takes 2 inches to stop, the average force on him is: Fav = (250lb)(60in/2in) = 7500lb But if he takes 1in to fall, the average force is 15000lb and if he takes 6in to fall, the average force is 2500lb. So the stretchiness of the rope matters a lot. It also depends on the surface the block is on. To take one extreme, if the block is on a completely slippery surface, it will eventually be dragged off the edge as well. You need to be more specific because the problem is underspecified. The OSHA regulations must make some assumptions about stopping distance (or stopping time) to be able to calculate this. Also, how does the block fit into the system? Is there a rope up from the man, through a pulley, and then down to the block, so that if the man falls the rope will be trying to lift the block into the air? Or does the rope go horizontally from the pulley to the block so that if the man falls, the rope will be trying to drag the block along the surface it's sitting on? If it's the latter, the OSHA regs must also make some assumptions about how rough/slippery the surface is. -- Rich Carreiro rlcarr@animato.arlington.ma.us posting-account=8jB0Vw0AAACJmu18OPPftv6igzbfKPoN 1) Deduct an equation of points in the plan,whose sum of the distances to the points (7,1) and (7,7) is 14. 2)Given the conics of equations 4y^2-^x^2=64 and y^2=-2x/3 write equations of the transformed associated with the vector (-5,-1).What are the focuses and the vertices of both and the equations of the directrix for the parabola? 3)Represent in the Argand plan 4)Determinate an equation of a hyperbola with asymptotes y=6x and y=-6x,knowing that the intersection of the conic with the axis of the ordenates are the points (0,9) and (0,-9) 4)Prove that the difference of the distances from the point (5/2,3) to the focuses of the hyperbola 4x^2-y^2=4 is equal to 4 5)Identify the lines: a)3x^2+4y^2-12x+12y+9=0 b)9x^2-4Y^2+18x+8y-31=0 c)x^2+6x+4y+9=0 4.1) y=6x and y=-6x V(0,+-9) b=9 a=3/2 y^2/81 + 4x^2/9 = 1 6) Say please when asking people to do ones homework. [eight problems] They're quite interesting; why don't you show us what you've done to work on them? The net, like God, helps those who help themselves. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ 2)The first eq should be 4y^2-x^2=64 Well,this is a nice exercise taken from another axiomatic matter of course that is a good bridge for linear algebra's analytic geometry.Oh,well i learn them with linear algebra,but they could be teached,also,as solid geometry along with the rest... Hope you get to the despair!! considering some of the mindbending equations and discussions I've witnessed on this board, this will probably seem like a piece of cake to all of you. I've got a differential equation that I can't quite pull an explicit solution out of and if someone could point me in the right direction of an explicit solution, I'd be eternally grateful. dy/dx = (3x^2 - e^x)/(2y-5) y(0) = 1 this is a separable equation, so.. Int( 2y-5 ) dy = Int( 3x^2 - e^x ) dx this leads to y^2 - 5y = x^3 - e^x + c [consolidated the two constants into one on the right side] if you solve for c, you get -3, so substituting that back in.. y^2 - 5y = x^3 - e^x - 3 and heres where I'm not sure of what happens next, I have tried solving for a y value ( y^2 - 5y + 3 = x^3 - e^x, and then using the quad forumula to get two values of y), but that method brings me to a dead end. Anyone got any suggestions as to what I should do next? -matt y = (5 +- sqr(25 - 4(3 + e^x - x^3)))/2 You don't want to go there, I see no point in it. You got an implicit equation for y that includes both branches while avoiding negative surds. So leave well enuf alone. Let's check it. 2yy' - 5y' = 3x^2 - e^x An instant push over. If you must have an *explicit* solution, you will just have to perservere with the quadratic formula; you are on the right track. If you do an implicit plot of your solution, you get a closed curve that indeed has a top and bottom half, but it passes through the line y = 5/2 where the derivative doesn't exist so only the bottom half is the solution through (0, 1). So if you do use the quadratic formula, you only want the - term in the formula. Personally, I would leave it in implicit form if that is allowed for your purposes. --Lynn Hello all, I just placed some pdf versions of the files. http://www.geocities.com/titus_piezas/ramanujan.html The latest one, On Ramanujan's Other Pi Formulas also uses Dedekind eta functions. --Titus posting-account=8jB0Vw0AAACJmu18OPPftv6igzbfKPoN sqrt(0)= 0 = log(1)= log(sqrt(1)). math1234 I've got three equations that I need some help to simplify. sqrt( (10-x)^2 + y^2) - sqrt(x^2 + y^2) = deltaAB sqrt( (10-y)^2 + x^2) - sqrt(x^2 + y^2) = deltaAD sqrt( (10-y)^2 + (10-x)^2) - sqrt(x^2 + y^2) = deltaAC deltaAB, deltaAC and deltaAD are all known values. I need to solve the equations for x and y but dont know how. Hoping for some adviceinput. May be the given system of equations is surdetermined: sqrt(x^2+y^2) has three different 'counterparts' or two unknown x,y and three RHS delta_ab,delta_ac,delta_ab. Try solving pairs ... Alain. I did not quite understand what you we're suggesting in your post. Could you please elaborate a wee bit. Please, can everybody help me to find the solutions x(t), y(t) of this differential system : a1,a2,a3,b1,b2,b3 are real constants. x'(t) = a1 x(t)*(1 + a2 x(t)+ a3 y(t)) y'(t) = b1 x(t)*(1 + b2 x(t)+ b3 y(t)) Merci beaucoup. Erratum x'(t) = a1 x(t)*(1 + a2 x(t)+ a3 y(t)) y'(t) = b1 y(t)*(1 + b2 x(t)+ b3 y(t)) Merci beaucoup. X-Proxy-User: $$rnw$j2sdg_ This is unprovable because it is false. If, one other hand, k and i are nonnegative integers, then it is sufficient to observe that [10^k-10^i] = (1)^k-(1)^i = 1-1 = 0 (mod 9). If modular arithmetic is not available to you as a method of proof in this case, then do double induction instead. First, prove by induction that 9|(10^k-1), and then prove that whenever 9|(10^k-10^m), then 9|(10^k-10^{m+1}). The are forced to consider positive integers only (say, in a foundations of mathematics class, where the proofs can get quite pedantic), a simple way to get around that is to do again show that whenever m is in A_k, then so is m+1. --- Stan Liou 1), with m = k-i. But for any x, x^m - 1 = (x-1)(x^(m-1) + x^(m-2) + ... + 1). So 10^m - 1 = (10 - 1)*integer and we're done. Since 10 is congruent to 1 (mod 9) 10 to any pos integer power is also congruent to 1 (mod 9) and, as noted by another post, 10^(k-1)-1 is a factor of the given expression. Responding to: I've been going in circles to prove this. Can anyone help? Let d=k-i, then your number M=(10^k - 10^i)=(10^i)*(10^d -1) and of course 10^d -1 is 9*(number 111...11 with d 1's). Are you sure it is? If i= -2 and k= -1, that is 1/100- 1/10= -9/100. I will assume you meant for POSITIVE integer values of k and i. Looks like a candidate for induction. Let i be any positive number and define n= k-i. Then the proposition becomes Let i be any positive integer, then (10^(i+n)- 10^i) is divisible by 9 for all positive integers n. If i is any positive integer and n=1, then 10^(i+n)- 10^i= 10^2- 10= 90= 9*10. Yes, it is true for n= 1. Now, assume 10^(i+k)- 10^i is divisible by 9 for some specific k. That is, 10^(i+k)- 10^i= 9m for some integer m. The 10^(i+(k+1))- 10^i= 10(10^(i+k))- 10(10^i)+10(10^i)- 10^i= 10(10^(i+k)- 10^i)+ (10-1)(10^i)= 10(9m)+ 9(10^i)= 9(10m+ 10^i), a multiple of 9. If 0 <= k,i then since 10^k = 1 mod 9 and -10^i = -1 mod 9 the sum is going to be equal to 0 mod 9.