Recent content by Jakob1

Thank you for all your help, Euge. I wonder if you could also take a look at my proof of the fact that $$||m_{xy}|| = d(x,y)$$. $$AE_0(X)$$ is a linear space and for a fixed $$x_0 \in X$$, $$m_{x x_0}, \ x \neq x_0$$, form its basis. Any real valued function $$f$$ on $$X$$ such that...

Hi. I've just solved a problem from functional analysis and I would be very glad if you checked if everything is all right: $$(X, d)$$ is a metric space, $$AE_0(X) = \{ u : X \rightarrow \mathbb{R} \ : \ u^{-1} (\mathbb{R} \setminus \{0 \} \ \ \text{is finite}, \ \sum_{x \in X} u(x)=0 \}$$...

Ok, I see that now. Now there remain only two integrals: $$\int_0^{\infty} e^{-x^2} dx = (1+i) \int_{0}^{\infty} e^{-2iy^2} \,d y = (1+i) \int_{0}^{\infty} (\cos 2y^2 - i \sin 2y^2) \,d y$$. $$= \int_{0}^{\infty} \cos 2x^2 \,d x - i \int_{0}^{\infty} \sin 2x^2 \,d x +i \int_{0}^{\infty} \cos...

I need to prove that this integral $$\int_{0}^{r} \frac{e^{y^2 - r^2}}{e^{2ir}}\,d y$$ vanishes as $r$ approaches infinity. $$| e^{y^2 - r^2}e^{-2ir} | \le | e^{y^2-r^2} | $$, because $$|e^{-2ir}| \le 1$$. But now I don't see how I can apply the de l"Hospital rule, because there is no denominator.

Should I use de l'Hospital's rule to the function I'm integrating? If so, I get $$\frac{-2re^{y^2-r^2}}{-2( \sin 2yr - i \cos 2yr)}$$. Did I misunderstand you? What about $\sin t^2$?

$$\int_0^r e^{y^2-r^2-2iry} dy = e^{-r^2} \int_0^r e^{y^2-2iry} dy $$ I thought I could estimate it somehow or use the error function (this integral is equal to $$\frac{\pi}{2}(erf(x)+erf(1+i)r)$$), but I haven't come up with anything so far.

I'm sorry for the confusion. Let me try one more time. The equation of the line passing through $r$ and $r+ir$ is $x=r$, so $z=r+iy$, (because $ z=x+iy$), so $dz=idy$ and $ z^2 = (r+iy)^2=r^2-y^2+2iry$, so the integral over the vertical segment is $$\int_0^r (e^{y^2-r^2-2iry})dy$$. Is it...

I've noticed my mistake and I've just edited my post. Could you tell me if it is all right now? On this interval $[r, r+ir]$ we have $z= (r + ir)y$ so $dz=(1+i)dy$ and $z^2 = (ry+iry)^2$, so here we get: $$ \int ^{r+ir}_r e^{-(ry+iry)^2}idy = \int ^{r+ir}_r e^{-2r^2y^2} i dy$$Could we use this...

I see. You've determined the equation of the line on which lie $$0$$ and $$r+ir$$.However, I still do not see how to use it to derive $ \sin t^2$ from this integral(s).

If we go this way: $$0 \rightarrow r \rightarrow r+ir$$ we get: for the segment $$[0, \ r] : \ z=x$$, so $dz = dx$ and $z^2 = x^2$ $$ [r, \ r+ir] : \ z= r + iry$$, so $dz=idy$ and $z^2 = (r+iry)^2$. So the integral has this form: $$\int_0^r e^{-x^2} dx + i \int_r^{r+ir} e^{-(r+iry)^2} dy$$...

Thank you. So I know that on the intervals $[0,r]$ the integrals $$\int^r_0 e^{-iz^2} \,dz = \int^r_0 e^{-it^2}\,dt = \int^r_0 \cos(t^2)\,dt-i\int^r_0 \sin(t^2)\,dt$$ are equal. But on this interval $[r, r+ir]$ we have $z= (r + ir)y$ so $dz=(1+i)dy$ and $z^2 = (ry+iry)^2$, so here we get: $$...

Hello. I'm having trouble calculating $\int_0 ^{\infty} \sin t^2$ using the fact that $\int _{\partial Tr} e^{-z^2} dz = 0$, where $Tr = conv (\{ 0, r, r + ir \})$ (a triangle). I'm aware that I need to somehow transform $e^{-z^2}$ to get $\sin t^2, \ \cos t^2$ but I don't know how to do...

Yes, I mean isomorphic as vector spaces. I'm sorry, my question wasn't clear. Actually, I need to show that there are two equivalent ways of introducing dot product on $$\Lambda_p (V^*)$$. Let $$\beta: V \ni v \rightarrow <v | \cdot> \in V^*$$ - here $$< \cdot | \cdot>$$ is the inner product...

Hello. I've just read about natural identifications of exterior powers with spaces of alternating maps, etc here: Some Natural Identifications However, I have problems showing that the following operations give the same space: $$V \rightarrow \Lambda_p V \rightarrow (\Lambda_p V)^* \cong...