Recent content by Despondent

Unfortunately with the transitition from Word 2003 to 2007, MS decided to be a nuisance and perform a complete overhaul of the interface. It doesn't have Equation 3.0 anymore, it's been replaced with Equation Tools which has a rather poor set of matrix options compared to it's predecessor...

Hey everyone. I can't figure out how to use "Equation Tools" to enter a matrix with dimensions of my choosing. I can only find a very limited set of default matrix sizes to choose from, such as 1 by 2, 3 by 3 and 4 by 4, which are not sufficient for my purposes. More specifically, I'm trying to...

Maths competitions like the AMC are not indicative of mathematical aptitude. Many people who have done extremely well in such competitions say that their results are due to them having done some work outside of the normal syllabus to help prepare for them. At this stage of your academic life...

The VCE is different to the system in Queensland. A brief outline of VCE in year 12: I'm sure you'll be given info by your school on how the ENTER is calculated from your subject scores so I'll skip over that. - Each raw subject score is out of 50 and is assigned to you based on your...

What are you trying to get at? If you mean banging your head in a literal sense then that's complete rubbish no matter how you look at it. Assuming that I'm not way off track in reading between the lines, it is very important to have determination but if you don't recognise your own limitations...

If you were working in polar coordinates for example, the 'obvious' thing to do would be to write \delta = \delta \left( r \right) but this is incorrect since the delta function would not satisfy all of the required properties. I can't remember exactly off the top of my head but in cylindrical...

The singularity at e^{\frac{{i\pi }}{4}} is a simple pole and the numerator of the integrand is non zero and holomorphic at the singularity. So you can calculate the residue by using the formula {\mathop{\rm Re}\nolimits} s\left( {\frac{1}{{z^{^4 } + 1}}} \right) = \frac{{1_{z = z_0 }...

Why don't you just calculate the residue directly?

The first one should be fairly easy. In the second one what you need to do is eliminate x so that you can solve for y in terms of u and v only and then eliminate y so that you can solve for x in terms of u and v only. If you're stuck on solving for x and y, then just consider u^2 + v^2 and u/v.

I think you should go with my suggestion to write the integrand in terms of the complex exponential. You don't need any tricks here (eg. multiplying the numerator and the denominator by some function). The contour C you should use has vertices -R, R, R + (pi)*i, -R + (pi)*i which you have...

A rectangular contour, as the thread starter has attempted to use, is the way to go. The method is to calculate the integral in two ways, one directly via the residue theorem and the other by parameterising the contour. A semi circular contour will probably not work. This is because as the...

\int\limits_0^\infty {\frac{{\cos x}}{{\cosh x}}dx} = \frac{1}{2}\int\limits_{ - \infty }^\infty {\frac{{\cos x}}{{\cosh x}}dx = \frac{1}{2}\int\limits_{ - \infty }^\infty {\frac{{e^{ix} }}{{\cosh x}}} } dx Doing it this way, you can avoid having to bound the cosine term explicitly...