Analysis and Calculus

i need to evaluate the follows integral: [math]\oint dy (A' e^{A})' [/math] where ' is derivation respect [math]y[/math] and [math]A(y)=A(y+2\pi)[/math] and too [math]A(y)'=A(y+2\pi)'[/math] where [math]y \in (-\pi,\pi)[/math] ia a angular coordinate , and [math]A' [/math] is discontinuous in [math]-\pi,0,\pi[/math] the result is zero, but i think that is NOT CORRECT to say: [math]\oint dy (A' e^{A})' = A' (\pi)e^{A(\pi)}- A' (-\pi)e^{A(-\pi)} =0 [/math], since [math]A' [/math] is discontinuous in -pi and pi

f:R-->Q show that there is a q \in Q such that f^-1(q) is infinite set in R ?????

Suppose we were given the following inequality :[math]\frac{x-2}{x}>x+1[/math] ,and the following two solutions were suggested: Solution No 1: for x>0 the inequality becomes (after multiplying across):[math]x-2> x^2 +x[/math] ,or [math] 2+x^2<0[/math],but since [math]2+x^2\geq 0[/math] we have a contradiction ,hence the inequality is satisfied for all x<0 Solution No 2:[math]\frac{x-2}{x}>x+1\Longleftrightarrow\frac{x-2}{x}-(x+1)>0[/math] . And working on the left hand side of the equation we have : [math]\frac{-2-x^2}{x}>0[/math] ,or [math]\frac{2+x^2}{x}<0[/math],and since [math]2+x^2\geq 0[/math] we conclude that x<0…

I was wondering how you integrate - with simply the integral of the power rule (n^x)' = nx^n-1, when the equation, such as (15x^3 + 8x^2 - 5/x) doesn't have any multiples for the antiderivative of 15x^3 or 8x^2... do you just leave them or something...

[math] ax^2 + bx + c = 0 [/math] is general form by using factorisation, we can turn it into this form let's say... [math] (x - r_1)(x + r_2) = 0 [/math] then we get the roots as follow [math] x - r_1 = 0 [/math] [math] x = r_1 [/math] & [math] x + r_2 = 0 [/math] [math] x = - r_2 [/math] compare to forming an equation out of roots(where roots are given) it's different story, where it's giving me some headache to think about... usually, teacher will give formula [math] x^2 - (S.O.R) + (P.O.R) [/math] S.O.R = sum of root P.O.R = product of root but, I personally don't like remembering and plugin stuff in ma…

Any suggest about how to solve this ecuation will be appretiated [math] x(\csc(x)+1)=\pi[/math] Thanks in advance.

How do I determine whether a triangle with a given vertices is a right triangle? It gives... (7,-1) (-3,5) (-12,-10)

This might be a dumb question but when solving ODE's with scilab and you set the independent variable points at which you want to evaluate, what is the middle number used for in t = 0:0.1:5 for the example: function dx = f ( t , x ) dx = sin (2 t ) ; endfunction t0 = 0 x0= -0.5 t=0:0.1:5; x = ode(x0, t0, t, f); plot2d(t, x) Here's the link for the example: http://scilab.in/files/workshops/15-04-2010-mumbai/sengupta-ode.pdf Thanks for the help!

prove that the following sequence converges: [math]x_{k+2} =\frac{2}{x_{k+1} + x_{k}} ,x_{1},x_{2}>0[/math]

I was solving some differentials for practice, and the final question I was given was [math] (x^2-1) \frac{dy}{dx} = xy [/math] I went about solving this by doing (some steps missed out): [math] \int\frac{1}{y}dy = \int x^3-x dx [/math] [math] ln(y) = \frac{2x^4-4x^2}{8}+k [/math] And then, I was left with this, extremely irritating answer, which didn't look right at all, and incidently, wasn't right: [math] y = Ae^\frac{2x^4-4x^2}{8} [/math] Note: A = ek I don't think I've done anything wrong with the integration, I just have a feeling there's a way to simplify this that I don't know. Any suggestions please: I need to learn! ADDITION: …

I'm not fantastic at integration, but I can integrate any number of polynomials pretty easily. I was given the equation "y=x(3-x)" to integrate. so I naturally treated it as a factorised quadratic, expanded it, and did: Int.(-x2 + 3x)dx = -1/3x3 + 3/2x2 I therefore assumed that: Int.(-x2 + 3x)dx = Int.(x(3 - x))dx It turns out i was wrong, and that I can't just expand these brackets, as the actual answer is (according to an integral calculator): -((2x3 - 9x2)/6) I need to know why I can't just expand these brackets and integrate, and how to integrate a factorised quadratic like this one. Please help me I also don't think a "u" substitu…

I'm doing a biology lab and was wondering, a basic version is this I can either put 0 1 or 2 of two types of organisms in a test tube and the test tube can then either be put into the dark or the light. I was wondering how many test tubes I will need to conduct this experiment. I'm thinking that you would multiply 3 by 3 by 2 because there are 3 possible number of each organism and their is 2 places they could be put. If someone could explain how to do this type of problem I would be very grateful. I don't actually need to fill all possibilities, just enough to answer some lab questions, which I have answered before I did anything. I was just wondering what the formula wo…

Hope this is in the right section. Having trouble ironing out an apparent inconsistency in matrix trace derivative rules. Two particular rules for matrix trace derivatives are [math]\frac{\partial}{\partial\mathbf{X}} Tr(\mathbf{X}^2\mathbf{A})=(\mathbf{X}\mathbf{A}+\mathbf{A}\mathbf{X})^T[/math] and [math] \frac{\partial}{\partial\mathbf{X}} Tr(\mathbf{X}\mathbf{A}\mathbf{X}^T)=\mathbf{X}\mathbf{A}^T+\mathbf{X}\mathbf{A}[/math] Now assume that [math]\mathbf{A}[/math] is diagonal and [math] \mathbf{X}[/math] is anti-symmetric. Then by the cyclic property of the trace, [math]-Tr(\mathbf{X}^2\mathbf{A})=Tr(\mathbf{X}\mathbf{A}\mathbf{X}^T)[/math]. So t…

My question is how to compute R(dx). But before I can ask that I have to write down the background to my problem, so bear withme ---------------------------------------------------------------------------- A tempered stable distribution is when a stable distribution is tempered by an exponential function of the form [latex]e^{-\theta{x}}[/latex]. In my particular case we are using a tempered stable law defined by Barndorff-Nielsen in the paper "modified stable processes" found here, http://economics.oul...nmsprocnew1.pdf. In Barndorff's paper, [latex]\theta = (1/2)\gamma^{1/\alpha}[/latex], hence the tempering function is defined as [latex]e^{-(1/2} \gamma^…

I'm struggling to work out how to integrate the following [latex]\int_0^t(\gamma^{1/\kappa}-i\zeta{w}(1-t/s)_+^{H-1/2})^{\kappa}ds[/latex] here (.)_+ denotes the positive part if I did not have the ^(H-1/2) I can do it, alas it does have it! and so it stumps me on how to evaluate this integral. any advice much appreciated

based on the example 2 in this website why does the normal equation doesn't perpendicular to the tangent? or i do any calculation wrong?

Why are P.D.E.s are solved without excitation and then applied on the solution with BC's? Is this general? What if the distribution was total-domain-space-functional? Just for analysis.

There was a post a while back on "1/0", and I was disappointed to see a post of mine deleted. So I'll be brief and repeat my previous remark. While 1/0 is generally undefined for fields, it is defined on the Riemann Sphere for the so-called extended complex numbers. On this complex manifold, 1/0 = [tex]\infty[/tex], the "point at infinity". Wolfram|Alpha and Mathematica implement this sort of complex arithmetic, albeit a bit inconsistently. Here's what Wolfram|Alpha thinks 1/0 is: http://www.wolframalpha.com/input/?i=1%2F0

While reviewing basic calculus, I noticed that the curve (1+t^2,t^2,t^3), which clearly has a cusp at (1,0,0), has a derivative curve (2t,2t,3t^2) which is clearly smooth. This struck me as odd since differentiation usually seems to turn cusps into discontinuities, whereas integration smoothes out a curve, especially a curve described by polynomials. In fact, in general I have always taken a curve to be smooth iff it has a continuous derivative, which this curve has, and yet a cusp cannot be smooth in any sensible sense. I suspect the explanation is relatively simple - just something I'm missing. Thx in advance.

I just learned that the normal distribution is the solution to [math] \frac{\mathrm{d}y}{\mathrm{d}x}+yx = 0 [/math] What problem led to this differential equation? How does this equation lead people to think of distributions? In a "real" world problem (physics) what could [math]y[/math] and [math]x[/math] be?

Question: Let A =(1+R)〖(-4+R^2)〗^2 B =〖(6+4R+B^2 R-5R^2-R^3+R^4+B(R^2+R-2))〗^2 C =(1+γ)〖(-4+γ^2)〗^2) D =〖(6+4γ+β^2 γ-5γ^2-γ^3+γ^4+β(γ^2+γ-2))〗^2 ) I need to prove that A /4B - C /9D > 0, given 0 < γ < R < 1, 0 < β < B < 1. How to do it? If I can use Mathematica to help, that will be great too. What I am thinking now is to find the Minimum of (A /4B - C /9D). If the minimum >0, then A /4B - C /9D >0. Is that a right direction to go? How can I write the input code for this including the constraints 0 < γ < R < 1, 0 < β < B < 1? Thank you so much for any hints or help!

before getting started... i know how to use discriminant for Quadratic Equation/Function which is when [math] b^2 - 4ac > 0 [/math] so, the root(x) would have 2 intersect point and [math] b^2 - 4ac = 0 [/math] so, the root would have 1/equal intersect point and [math] b^2 - 4ac < 0 [/math] so, there are no root. but, I can't get the relation between those.. I mean, why [math] b^2?[/math], why [math]-4?[/math] why a and c?, why > or = or < ??? http://www.coolmath....arabolas-01.htm this website give me good concept of understanding the Quadratic Equation/Function in General Form/Vertex Form, but I found none for 'discriminant' w…

Okay, in the Escape Velocity Equation, with the usual notations, v2=2gR2/r + (v02-2gR) a few articles like www.math.binghamton.edu/erik/teaching/02-separable.pdf and http://www.uv.es/EBR...5000_17_84.htmlgive the following explanation: ... ... ...(implying that v0>=√2gR) That raises a few doubts in my mind: (1)Does the term 2gR2/r become zero? (2) If the answer to (1) is yes(i.e. 2gR2/r=0), then how come v0 is allowed a value equal to √2gR?(since, substituting v0=√2gR in the original equation(i.e. v2=2gR2/r + (v02-2gR)) will yield v=0(which is obviously not desired if a particle has to escape the Earth) (3)If the answer to (1) is no(i.e. 2gR…

I'm trying to solve this function. [math] = \frac{1 + (\frac {2x + 3}{3x- 4})}{(\frac {4x + 6}{3x-4}) - 3} [/math] [math] = \frac{\frac {2x + 3 + 3x -4}{3x- 4}}{\frac {4x + 6 - 9x +12}{3x-4}} [/math] [math] = \frac {5x - 1}{3x- 4} / \frac {-5x + 18}{3x-4}[/math] [math] = \frac {5x - 1}{3x- 4} * \frac {3x-4}{-5x + 18}[/math] [math] = \frac {5x - 1}{1} * \frac {1}{-5x + 18}[/math] [math] = \frac {5x - 1}{-5x + 18}[/math] but how come mathematica got this? mathematica 1 mathematica 2 where does the " -1 " came from???

Can someone explain to me (in layman terms, mostly) what "b", "p", and "beta" stand for in Table 2 in this article: http://www.plosone.o...al.pone.0017006 Also, I have trouble reading parts of the Tables (especially Table 2) because I don't have any sense of what "units" are being used for these values. In Table 1, mL is specified in one case, so that makes sense. In other sections it's simply the number of people in the study, but how do you read the numbers that don't have units, like the "Stress" or "Social support" values, or anything in Table 2? Thanks.