Recent content by Jory

You can simply substitute x^3 in for x, but remember that the radius of convergence needs to be taken into account. I.e, if f(x) converges for |x| < 2, then f(x^3) would converge for |x^3| < 2. Try it out though. You can compute the Taylor series from the definition, and then by substituting...

Hmm I'm still not getting the correct answer. Using the Taylor expansion of e^ix, then dividing by x^2, gives me a coefficient of -i in front of the z^-1 term does it not? Which makes the Residue = 2Pi, then dividing this by 2 gives me Pi, not Pi/2 as the answer. Where am I going wrong?

Awesome Thanks bud

Just to point out, that wasn't the OP ;), but yeah that helps. Will take another look Edit: So can you calculate the residue at 0 using the Laurent expansion of e^ix, which gives the answer as Pi? Thanks

I think this definition stated here looks a bit like Uniform continuity, which has a subtle but important difference. In 'normal' continuity, delta can depend on x, so f(x) : R -> R, f(x) = x^2, is continuous With Uniform continuity, delta cannot depend on x, so is a global property of a...

Homework Statement Evaluate \int \frac{1-cos(x)}{x^2} dx given that it exists (The integral is between 0 and Inf, couldn't figure out how to add limits) Homework Equations No idea, looking round, it seems this is related to the Dirichlet Integral but the proof is way beyond the level I'm...

Of course, thanks

It should be mentioned when you solved basic PDEs. For example if U is a function of (t,x) Ut = f(t,x) You would integrate wrt t, and add an arbitrary function of x. When I was taught this, he simply called them arbitrary functions. I'm not sure if they have a special name.

C(alpha) is a function of alpha only, so it's derivative with respect to t = 0 So instead of adding a constant, you add an arbitrary function of the non integrating variables.

Well alpha is presumably the variable you use for the Fourier transform of your function, so it is to say, since when solving the resulting ODE you integrated the RHS with respect to t, it is entirely possible that the standard integration constant is instead a function of alpha. (Since d/dt...

\partial U=-k^{2}\alpha^{2}Udt Here, the terms concerning U need to be brought on to the left hand side, and then both sides integrated. You will end up with x as a function of ln(y), then exponentiating both sides will give you the form you have

Homework Statement let \Gamma \subset C be the circle of radius 1 centred at 0. Let f: C \rightarrow C be an entire function such that for every z \in \Gamma f(z) = z Show that f(z) = z also on Int( \Gamma ) Homework Equations (f(z) - f(z0) )/(z - z0 ) = f'(z0) perhaps? The...