Recent content by BrainHurts

I don't remember, but in looking at your argument, why don't you include [\itex]|x|\geq 0[\itex]. Secondly, if you're going to do it for x<0, use the definition of |\cdot|, do the same thing, multiply both top and bottom by the conjugate and I think you have what you got. I hope this helps.

And the sample is hit with radiation and the radiation goes through the sample to which this radiation goes through the same Michelson Interferometer path?

I wish I knew how to draw a diagram. Let me see if I understand what you're saying. So the radiation goes through the interferometer first (as a reference), then the radiation goes through the sample. Do you then compare intensity values of the beam no sample to the intensity values of the beam...

I've been reading up on Fourier Transform Infrared Spectroscopy and the Michelson Interferometer. My main sources are "Principles of Instrumental Analysis" by Skoog etc and Fourier - Transform Infrared Spectrometry by Griffiths and Haseth. I believe I understand the theoretical principles...

I see, very cool, I mean I always felt the "speed of convergence thing" but that does not really stick to my head. It definitely will now. I definitely like the numerical analysis as opposed to analysis proper mention. That definitely cleared the air a bit. I don't know if I felt the water...

I really like this example, but I was thinking in the sense of a more physics like interpretation if that makes any sense.

Yes, but I think Erland hit that point, which makes more sense. I totally feel you on your "not quite proof". Do you have a physical meaning for uniform convergence vs. pointwise convergence by any chance?

That makes so much more sense Erland. I didn't look at it that way.

So I'm reading "An Introduction to Wavelet Analysis" by David F. Walnut and it's saying that the following sequence " (x^n)_{n\in \mathbb{N}} converges uniformly to zero on [-\alpha, \alpha] for all 0 < \alpha < 1 but does not converge uniformly to zero on (-1, 1) " My problem is that isn't...

Thanks so much! I know we want to do that in order to make the linearization work correct?

So I'm reading the Example on page 161 of Differential Equations, Dynamical Systems and an Introduction to Chaos by Hirsh, Smale, and Devaney. I'm not understanding everything. So given the system x' = x + y^2 y' = -y we see this is non-linear. I get it that near the origin, y^2 tends to...

Yeah that looks better!

Thanks Mark, shortly after I saw that I did the work and Now I'm convinced x' = ax and y' = ay is now justified. Now suppose we chose the principal value log, will that then justify the separation of variables? I feel like what I'm about to do is a little naive, but let's give it a shot...

By directly do you mean separation of variables? \dfrac{dz}{z} = a dt . First we need to define complex integration which would require line integration. I think that's what you mean right? Thing is, antiderivatives are a little more trickier in \mathbb{C} than they are with straight up real...

So z'(t) = x'(t) + i y'(t), I just want to make sure that what I'm going to do is OK. I'm trying to solve z' = az where a = \alpha + i \beta z'(t) = az(t) \Rightarrow x'(t) + iy'(t) = a(x(t) + iy(t)) \Rightarrow x'(t) = a x(t), y'(t) = a y(t) If you could give me a justification that'll...