Recent content by sam2

  1. S

    C/C++ Using Complex C++ Class Header <complex>

    Hi, Has anyone here used the complex C++ class before Header <complex>? I am trying to do something VERY straightforward but there doesn't seem to be any way to do it! Basically, I define a complex, and then want to re-asign its real and imaginary parts: complex<double> A(1,1); //...
  2. S

    Fourier Transform of Probability distribution

    Hi, Sorry about the text, but Latex doesn't work. Can anyone please give me an outline for the derivation of the probability function by inverting its Fourier transform, i.e. P(X>x) = \frac{1}{2} + \frac{1}{\pi} \int_{0}^{\infty} Re \bigg[\frac{e^{-i \theta x}f(\theta)}{i \theta} \bigg]...
  3. S

    Complex Integral: Solving from Ln to ArcTan

    much appreciated
  4. S

    Complex Integral: Solving from Ln to ArcTan

    Hi George, Thanks for the reply but I am still stuck when I use your method. It makes perfect sense but I can't see where I am going wrong. I am doing the integral that I mentioned in my first post; but basically end up with the integral equal to tan^{-1} [ -2x/(1-x^2) ] Sorry, latex doesn't...
  5. S

    Complex Integral: Solving from Ln to ArcTan

    Gotcha. Many thanks.
  6. S

    Complex Integral: Solving from Ln to ArcTan

    Hi, I'm doing the following as an exercise to try and get my head around complex numbers. Specifically, I need to understand what it means to take the natural log of a complex number and what it involves. Say I wanted to integrate 1/ (1 +x^2) dx I know this is arcTan(x). I can also...
  7. S

    Solving Complex ODEs: Tips and Resources

    So do I just solve the ODE on the basis of the sign of the real part of the complax coefficient? I.e. I still don't know whether to use a tan or ln solution to the above. My b is indeed positive but I was worried about using a tan substitution because of the 'i'. So you are effectively saying...
  8. S

    Solving Complex ODEs: Tips and Resources

    Hi, Thanks for the replies. But I am still unsure as to why the solution is a tangent rather than a logarithm. We can rewrite the quadratic term in the ode as the difference of two squares, say: 1/a . df / (F^2 + D^2) = dx If D is positive then I agree that the solution uses a tan...
  9. S

    Solving Complex ODEs: Tips and Resources

    Hi, I have never had to handle ODEs where the coefficients are complex. Just wondering if solving this is even possible and whether you can point me to any sources/books. Say I had the ODE (df/dx) + a.f^2 + (b+i)f + c = 0 where f(x) is a function of x, a, b and c are constants, and i...
  10. S

    Is exp(i.x) with x=-infinity equal to zero?

    Got it. Many thanks for the replies. Regards, Sam
  11. S

    Is exp(i.x) with x=-infinity equal to zero?

    Hi all, Looking for some help on the following problem. Any replies much appreciated. I have the complex number exp(i.x) If x = - infinity, is this zero?? Is there any intuitive/straightforward value that it should be? I decomposed the expression into cos and sin and it looks like...
  12. S

    What Are the Hottest and Coldest Points on the Unit Ball?

    Hi, thanks for the replies... Integral I like your intuition... regards
  13. S

    What Are the Hottest and Coldest Points on the Unit Ball?

    Hi all, Came across this problem, but it has stumped me: Let S denote the unit sphere x^2 + y^2 + z^2 = 1, and let u = x + 2y + 3z be temperature at points everywhere in 3-space. Find the hottest and coldest points on the unit ball x^2 + y^2 + z^2 <= 1 I figured out that...
  14. S

    Solve Limit Question: f(x)=(1+.01x)^(10/x)

    How about using the Binomial Exapnsion to re-write the expression and then looking at whether you can simplify it when x--> 0?
  15. S

    Solving a Challenging Probability Problem - Detailed Steps

    What about using the Central Limit theorem? X(1) = The outcome from dice number 1, it can be 1,2...6 We want Y = X(1)+X(2)+... X(20). This should be approximately normal in distributoin. Mean = 20 *3.5, variance = 20*2.92. We want P(Y>100) Sam
Back
Top