Recent content by Semo727

I really heard this in context of ZF set theory... but maybe it was just some mistake in that text. Now I will maybe think about that this way until I learn more about logic and set theory :)

I have read, that properties of sets such as that every subset has supremum or that set is well ordered cannot be expressed in the language of first-order logic. Well, when I tried to write these things, I seemed to write them in first order language, which really bothers me. So please, tell me...

oh, there is a mistake in mine (and also HallsofIvy's) post. L' should be L'=\frac{d}{dx}(\frac{ae^{(b/a)x}d}{dx})+ce^{(b/a)x} and so the derivative "jump" should be probably \frac{e^{(b/a)x}}{ae^{(b/a)x}}=1/a which is also what one can guess also from the original linear operator...

Oh, yes, thanks. OK, so we can transfer the linear operator to desired form, but they are not the same. I mean, that L=a\frac{d^2}{dx}+b\frac{d}{dx}+c\neq L'=\frac{d}{dx}(e^{(b/a)x}\frac{d}{dx})+ce^{(b/a)x}=Le^{(b/a)x} So I think that also right side of the equation: \delta(x-y) should be...

Hello! I have problem with my homework, but what I'm going to ask you is not homework problem so I hope it is OK I'm writing it here :) I need to find Green's function for differential operator L=a\frac{d^2}{dx^2}+b\frac{d}{dx}+c i.e. find solution for differential equation equation...

logarithm is defined also for complex numbers. ln(z)=ln(abs(z))+i*arg(z), where z is complex number, abs(z) is complex norm of complex number z, and arg(z) is its argument. So if -1 is treated as complex number -1+0*i, expression ln(-1) gives sense, but the identity a*ln(z)=ln(z^a) is no...

And I also guess that (\vec{A}\cdot\vec\nabla)\, \vec{B}\neq\vec{A}\,(\vec\nabla\cdot\vec{B}) since \vec{A}\,(\vec\nabla\cdot\vec{B})=\vec{A}\,\frac{\partial B_x}{\partial x}+ \vec{A}\,\frac{\partial B_y}{\partial y}+\vec{A}\,\frac{\partial B_z}{\partial z}

How can vector multiplyed by scalar give scalar??

I have been talking about slovakia and czechia which might have looked like I was talking in general, BUT not at all. Yes, we have rather good math oriented high school classes and some competitions here (but this occurs everywhere I think), but still it is only like a drop in the ocean. My...

You should see some of my peers who stick their nose into things like groups or QFT. That's not normal anyway :)

oh, thanx, I don't know if I'm really proud, but I'm happy that I have learned some basics of calculus and abstract algebra so far and now, I can understand more interesting parts of physics, which a like most. For instance partial derivatives and variations are nothing difficult to understand...

Oh, thanks arildno! that was something I wanted to see :) definitely much nicer than what I have done. I was also thinking about tangent vectors d\vec{t}_{x}=\left(\vec{i}+\frac{\partial{f}}{\partial{ x}}\vec{k}\right)dx d\vec{t}_{y}=\left(\vec{j}+\frac{\partial{f}}{\partial{...

Surface Area (help me to prove something:) I was studying a bit about multiple integrals and found this theorem: If we have function z=f(x,y) which is defined over the region R, surface S over the region is S=\iint_R\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial...

:rolleyes: OK, thanks. I just thought that there is some much shorter way I don't know about becouse I don't know much about integrating methods. (I know just per partes and substitution)

Hello! I would like to count (see the way how to count) this integral \int_0^v \frac{1}{(1-v^2)^{3/2}} \,dv It should be \frac{v}{\sqrt{1-v^2}}. I have managed to count it (I have just derived the result, and followed steps in reversed order), but this method was a little bit clumsy, I...