Recent content by tohauz

I see it now, thanks

Appreciate your help, you i "proved" my first claim. Could you help to find my error? So, if for given x f(x) \leq a, then for \forall \varepsilon>0, \forall x \exists k such that \forall n\geq k f(x) - \varepsilon < f_n(x) < f(x) + \varepsilon \leq a + \varepsilon Since \varepsilon...

Hi, 1)is this true? If f_n(x) -> f(x) pointwise, then {x:f(x)<=a} = union{k=1^infty}intersection{n=k^infty}{x:f_n(x)<=a}. 2)if A is measurable set, subset of reals, then is A-const set measurable? Thanks

How can we negate using similar steps? $\forall a,b \in R$ with $a<b$ there is an $r \in Q$ with $a<r<b$ I do know the negation, but I need explain to smb who is learning this. Thanks

OK, I got the first one, hopefully somebody can help me 2nd. 1) Answer is 2n. Use <u,v>=|u|*|v|*cosa, it is nonpositive if a>=90 degrees. So in R^3 take i,j,k,-i,-j,-k. If you want to squeeze in other vector the angle between that and 3 of those is less than 90.

1. Suppose v_1, v_2, . . . v_k are non-zero vectors is R^n such that (v_i,v_j)<= 0 for all i,j. Determine, with proof, the maximal possible k for n = 3, and also for arbitrary n. 2. A is 2x2, A(1,1)=A(2,2)=x-1, A(1,2)=1, A(2,1)=0. Find invertible P,Q such that P*A*Q is diagonal. I tried...

hi. Spse you want to find limf(x,y) as (x,y)->0. You can use polar coordinates and get limf(rcost,rsint) as r->0. And these limits are same. Now, i initially thought that 1)if limf(rcost,rsint)is independent of t,then limit exists and is equal tolimf(x,y) as (x,y)->0. 2)If it depends on...

Could you please tell me how you got it?

I see what you are saying, but at least did I set it up correctly?

OK. Where is my mistake? Thanks i'm having trouble with finding it

That doesn't make a sense. Because, if a_{i}=r for all i's, we don't get the surface area ball with radiuis r.

Actually, I used same idea to find the volume and I got it: it is a_{1}*...a_{N}*meas(unit ball)

Suppose that ellipsoid has axis a_{1},...,a_{N}. Then S=\int_{{\sum\frac{x^{2}_{i}}{a^{2}_{i}}<1}dS(x)-surface integral. Then i solved for x_{N}=a_{N}(\sqrt{1-(\frac{x_{1}}{a_{1}})^{2}-...-(\frac{x_{N-1}}{a_{N-1}})^{2}}) and used that formula for evaluating the surface area, where you need...

I tried. But I'm getting something which is not easily integrable

Hi guys. What is [SIZE="5"]surface area of N dimensional ellipsoid? Any help is really appreciated.