What are bessel functions?
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They are solutions to Bessell's differential equation:\n\nx^2y''+xy'+(x^2-n^2)y=0.\n\nThe solutions of this equation are called Bessell functions of order n. Using standard techniques, it is possible to find power series solutions to this equation for every n and two independent power series solutions if n is not an integer.\n\nBessell functions come up naturally from the partial differential equations of physics and have applications ranging from electromagnetism to sound. The functions are oscillatory in form, but unlike trigonometric functions, they decay towards infinity. There are books written just about these functions and their many properties.
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intgral(from - infinity to infinity) of e^(-x^2)= sqrt(pi)\n\nThis is the area under the bell curve. It comes up very often in statistics.
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http://mathforum.org/library/drmath/sets/high_calculus.html\n\ngood luck...
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I believe it is a bilinear operator  \n\nhttp://en.wikipedia.org/wiki/Bilinear_operator\n\neg. used in Tensor product\n\nhttp://en.wikipedia.org/wiki/Tensor_product
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In mathematics, the hyperbolic functions are analogs of the ordinary trigonometric, or circular, functions. The basic hyperbolic functions are the hyperbolic sine "sinh", and the hyperbolic cosine "cosh", from which are derived the hyperbolic tangent "tanh", etc., in analogy to the derived trigonometric functions. The inverse functions are the inverse hyperbolic sine "arsinh" (also called "arсsinh" or "asinh") and so on.\n\nJust as the points (cos t, sin t) define a circle, the points (cosh t, sinh t) define the right half of the equilateral hyperbola. Hyperbolic functions are also useful because they occur in the solutions of some simple linear differential equations, notably that defining the shape of a hanging cable, the catenary.\n\nThe hyperbolic functions take a real value for real argument called a hyperbolic angle. In complex analysis, they are simply algebraic functions of exponentials, and so are entire.
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Try http://integrals.wolfram.com/
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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.
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An analytical function is a complex differentiable at every point in the domain.\n\nIt is most commonly used in Advanced Algebra (a.k.a. Algebra 2), Pre Calculus, and Calculus.\n\nIt is a type of domain that is used within a graphing calculator and is used to make certain calculations throughout most chapters of those three books.\n\nHope my Answer helped you out.
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KE = mv\nWrong: energy <> momentum\n\n2y = (2 m)cos(kx)\nRight: a lenght =  lenghth x a number between -1 and +1\n\nvf = vi + ax\nWrong: x should be t\n\ndensity = volume/mass\nWrong: fraction should be inverse
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M(t), the moment gerenating function associated to a pdf (probability distribution funciton) is given by\nm(t)=E(e^(tX))= Integral( e^(tx) f(x) dx) from -inf to + inf.\nif it exists, them m(t) is continuously diffrentiable in a neighborhood of the origin, and \nthe nth derivative of m(t), evaluated at t=0 is equal to the nth moment; that is, to E(X^n).
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http://en.wikipedia.org/wiki/Eigen_value
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Classes of Solutions to Mathieu's Equation\n\n  Floquet's theorem asserts that any solution to the Mathieu equation (Eq. gif) is of the form:[4,29,28]\n\n \n\nwhere is either real or imaginary and is a -periodic functiongif of . This assertion implies that solutions of Eq. gif can be separated into the product of a term of the form and a series of harmonics, , each of which is -periodic in . For values of with a non-zero imaginary component and , Eq. gif indicates that:\n\nThus, , yields an unbounded solution to the Mathieu equation. On the other hand, values of which are purely real yield bounded, oscillatory solutions for .\n\nThere are two types of bounded solutions to Mathieu's equation: functions of fractional order and functions of integral order. Functions of fractional order are obtained for . The oscillatory periods of the frequencies associated with fractional-order Mathieu functions are non-integer multiples of . The special case of corresponds to an -periodic (in terms of ) solution. These -periodic solutions are Mathieu functions of integral order.\n\nAll bounded solutions to the Mathieu equation--- those of fractional as well as integral order--- are described (in accordance with Floquet theory) by an infinite series of harmonic oscillations whose amplitudes decrease with increasing frequency:\n\n \n\nThe argument of in Eq. gif represents harmonics of frequency , while the term represents a (secular) frequency which is a function of the trap specifications (or, equivalently, of the Mathieu parameters a, q). The Mathieu parameters and also determine the coefficients , which approach zero as n increases.[36]
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http://www.math.niu.edu/~beachy/aaol/frames_index.html
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Integral of e^(x^2)dx = (Sqrt[Pi]*Erfi[x])/2\nErfi[x] = imaginary error function\nErf[x] is the integral of the Gaussian distribution, given by (2/sqrt(pi)) * Integral from 0 to x of (e^(-t^2))dt
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Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
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You use derivatives to compute how values are changing. A good example is to compute velocities.\n\nIntegration is used to compute areas and volumes.
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u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.