Recent content by Karamata

How to deduce is it \{\cos(\sqrt{2}t)(2+\cos t), \sin(\sqrt{2}t)(2+\cos t),\sin t \mid t \in \mathbb{R}\} submanifold of \mathbb{R}^3? First, this is curve, so I was thinking to find point in which this curve has two intersect, and then some neighborhood of that point isn't homeomorphic to...

This isn't good, because \alpha l(u) + \beta l(v)+c = \alpha \left(\displaystyle\int\limits_a^b f(x)udx+c\right)+\beta \left(\displaystyle\int\limits_a^b f(x)vdx+c\right)+c = \alpha\displaystyle\int\limits_a^b f(x)udx + \alpha c + \beta \displaystyle\int\limits_a^b f(x)vdx + \beta c + c \neq...

Nonononononono! First of all l(v)=\displaystyle\int\limits_a^b f(x)vdx+c, NOT l(v)=\displaystyle\int\limits_a^b g(x)vdx+d. Who give you rights to say g(x) or d? NOBODY! l(\alpha u + \beta v) = \displaystyle\int\limits_a^b f(x)(\alpha u + \beta v)dx + c, not +\alpha c + \beta c

Maybe you can prove it like standard proof of Cauchy–Schwarz inequality: 0 \le <x+\lambda y, x+\lambda y> = <x,x> + 2\lambda <x,y> + \lambda^2 <y,y>, and then choosing that \lambda = -\dfrac{<x,y>}{<y,y>} you will get |<x,y>|^2 \le <x,x> <y,y> So, maybe, but maybe, you can use \lambda =...

For n=0 you have \dfrac{1}{\pi}\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi} \cos (0 \cdot x) dx = \dfrac{1}{\pi}\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi} 1 dx = \dfrac{1}{\pi} \cdot \dfrac{\pi}{2} Edit: Oh, Infinitum was faster :) Sorry

Hi robertjford80! I deleted my post because there was error in him (oh, bad English) But, look at picture. They said \int_0^6 \int_0^y x \mbox{d}x\mbox{d}y, that is yellow region (x from 0 (parallel y-axes) to x=y, y from 0 to 6). r is moving from r=0 to y=6, so, y= 6= r \sin \theta...

Hi :) Can you recommend me some books about mathematics, but more philosophical focus maybe (i didn't read this book, so I don't know) Sorry for bad English

Yes.

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I think that best solution (for me) is @gopher_p solution. You know that: \frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+4}} + \cdots + \frac{1}{\sqrt{n^2+2n}}\le \frac{1}{\sqrt{n^2+2}} + \frac{1}{\sqrt{n^2+2}} + \cdots + \frac{1}{\sqrt{n^2+2}} = n\cdot \frac{1}{\sqrt{n^2+2}}...

Darboux sum is important here. We define upper Darboux sum and lower Darboux sum, and say that function is Riemann-integrable iff \displaystyle\sup_{P}L_{f,P} = \displaystyle\inf_{P}U_{f,P} iff \forall \varepsilon>0 ~\forall P ~ \exists \delta>0~ \mbox{if} ~\lambda(P)<\delta ~\mbox{then}...

See definition

Well, you can transponse left and right sides of D=P^{-1}AP, notice that D = D^T

It looks like that for k=0 they use l'hopital's rule, because when k=0 you have 0/0 (but maybe this is not the limit). I don't know what is k,N,j, etc.

Because of my bad English, I will show how to do this problem: Find the surface area of the portion of the cylinder x^2+z^2=b^2 lying inside the sphere x^2+y^2+z^2=a^2 where 0<b<a. Solution: P=2P(S_1)=2\iint_{S_1}^{}\mbox{d}S=2\iint_{D}^{} \sqrt{1+(y_x)^2+(y_z)^2}\mbox{d}x\mbox{d}z where...