cauchy  problem      ?
348048
A sequence {x_n} is Cauchy if given e>0 we can find an integer N such that || x_m - x_n || < e for all m,n > N.  I want to prove a lemma first:\n\nLEMMA: Let f: A -> R^m be uniformly continuous.  If {x_n} is a Cauchy sequence, then so is { f(x_n) }.\n\nProof: Let e > 0 be given.  We must show that there exists an integer N such that || f(x_m) - f(x_n) || < e whenever m, n > N.  Since f is uniformly continuous, we can find d > 0 such that || f(x) - f(y) || < e whenever || x -y || < d.  Since d is given -- and only depends on e! -- there is an integer N such that || x_m - x_m || < d for all m, n > N.  This means || f(x_m) - f(x_n) || < e for all m, n > N.\n\nWe now prove the main problem with the following.\n\nPROPOSITION: A sequence {x_n} is Cauchy if and only if lim x_n exists.\n\nProof: First assume that L = lim x_n exists.  Given e > 0, there exists an integer N such that || x_n - L || < e/2 whenever n > N.  For m, n > N, we have, by the Triangle Inequality,\n\n|| x_m - x_n || <= || x_m - L || + || x_n - L || < (e/2) + (e/2) = e.\n\nConversely, say that {x_n} is a Cauchy sequence.  We give a Proof by Contradiction, so assume that lim x_n does *not* exist.  That means given any point L in R^m, real e > 0, and integer N we can find an integer m > N such that || x_m - L || > e.  Since e is given, we can find an integer N such that || x_m - x_n || < e for all m, n > N.  Choose n = N+1 and L = x_n.  Then there exists an integer m > N such that\ne < || x_m - L || = || x_m - x_n || < e\n\nwhich is a contradiction.  Hence lim x_n must exist.\n\nCOROLLARY: If {x_n} is Cauchy and f: A -> R^m is uniformly continuous, then lim f(x_n) exists.\n\nProof: If {x_n} is Cauchy and f: A -> R^m is uniformly continuous, then { f(x_n) } is also Cauchy.  The limit of a Cauchy sequence exists, so L = lim f(x_n) exists.
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
427860
There is an infinite number of solutions, \nthe set of solutions is :\n{(k,-k+12):k belongs to the set of real numbers}
939642
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
483089
First write your equation using hyperbolic function:\n\ny = a cosh(x/a)\n\nDifferentiate, dy/dx = sinh(x/a)\n1 + (dy/dx)^2 = 1 + sinh^2(x/a) = cosh^2(x/a)\nSQRT (1 + (dy/dx)^2 ) = cosh(x/a)\n\nIntegrate from x = 0 to x = a:\n\nAnswer = [asinh(x/a)]@x=a  - [asinh(x/a)]@x=0\n       = asinh1 - asinh0\n       = asinh1\n       = a(e - (1/e))/2\n\nRefer to the following website for details on the catenary
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
817419
Proof.\n\nsin : [a, b] → ℝ is a continuous differentiable function, so by the mean value theorem, ∃ c ∈ (a, b) such that\n\nsin'(c) = (sin(b) - sin(a)) / (b - a), i.e.\ncos(c) * (b - a) = sin(a) - sin(b).\n\nNow since ∀ x ∈ ℝ, -1 ≤ cos(x) ≤ 1, it follows that\n\n| sin(a) - sin(b) |\n= | sin(b) - sin(a) |\n= | cos(c) * (b - a) |\n= | cos(c) | * | b - a |\n≤ 1 * | b - a |\n= | a - b |. ∎
950482
OK, Drummanmatthew.  It's been a long time since I did this but I think I remember enough to help you out.\n\nFrom what I remember, the potential, phi, is a function of x, y and z defined so that F = -grad(phi)  (I'm not clever enough to do symbols, I'm afraid) where grad is the operator (d/dx, d/dy, d/dz) where the d's are actually partial differential operators, rather than ordinary differential operators.\n\nIn other words, we need a function which when you diff wrt x gives you -2xy, when you diff wrt y gives -x^2-z^2 and when you diff wrtz gives you -2yz.  To me, it is the y part of this that gives it away.  It clearly has to be something plus something.  You need the bit with the z to differentiate away when you diff wrt x.  It seems to me that -yz^2 will do the trick.  Diff wrt x and it goes, diff wrt y and you get -z^2.  Diff wrt z and you get -2yz, as required.  To get the rest of it right you also need a -xy^2 by similar logic.\n\nSo, I reckon that phi = -y(x^2 + z^2) + c\n\nThe constant c should be obtainable from a boundary condition somewhere (I seem to recall that we often want phi to go to 0 as we go off to infinity but this is clearly not possible here).\n\nI'm not absolutely certain about the - but I think this correct apart from that.\n\nHope this helps.
510279
Jimbo, you gave up too soon......\n\nV = L*W*H = L*W*x in this problem\n\nV = (6-2x)*(4-2x)*x\n\nV  oila!
576554
Hint: If an analytic function is 0 on a set with its accumulation point, then the function is identically 0.\n\nNow, either f or g has to have this property since at each point one or the other is 0.
432536
Oops………..!\n\ndac2chari\n\n\n u = 1/(ln t) Then du = t dt ………………..?  How that come …!\n\nIf u = 1/(ln t) then  u = [ ln t ] ^-1 \n\nAnd  du = - [ln t ] ^-2 [ 1/t ] dt = dt / [-t (ln t ) ^2] … please be careful.\n------------------------------------------------------------------------------------\n\nOK… this integral can’t carry out using analytical method; however, the flowing one can be solve\n\n∫ dt / [t ln(t)] \n\nu = ln (t)\ndu =  dt / t  \n\nGives\n\n∫ dt / [t ln(t)]  = ∫ du/ u  ln(u)\n\nSubstitute for u\n\n∫ dt / [t ln(t)]  = ln [ ln (t) ]
735282
Seems like you are missing some parentheses and an exponent. Please check to make sure yiu have copied the problem correctly and resubmit.
970181
1) This has no elementary solution\n2) This one, however, does have a solution, with the substitution e^x = tanu.  You are at the answer arctan(e^x)\n\nSteve
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
260164
If |z|<1, then z^n converges to 0. If |z|>1, then z^n diverges since |z|^n goes to infinity. For |z|=1, |z^n -z^(n+1)|=|1-z|, so z^n can't be Cauchy unless z=1. Thus, {z^n} converges if and only if |z|<1 or z=1.
277316
let |f|=k (some constant).let u,v be the real & img parts of 'f'.\n            the we have |f|=u^2+v^2, u^2+v^2=k^2\n      differentiating partially,wrt to x&y ,and \n\n    USING CAUCHY-REI EQ'S FROM TWO RESULTING EQ'S \n\nu finally get     (u^2+v^2)(du/dx)=0.\n                 &  (u^2+v^2)(du/dy)=0. \nfrom above two eq's   we have \n                     k^2=u^2+v^2=0  implies u=0&v=0\n\n                  so 'f' is constant.