Recent content by kekido

Ok, here "boundary" is the set of all boundary points, i.e., \partial(Q)=R You're right, the interval should be like what you said. I was being sloppy here. Since the boundary point is defined as for every neighbourhood of the point, it contains both points in S and S^c, so here every...

I don't see it as a complication. Rather, it's the definition of an even number being able to be divided by 2 with no remainder, hence 2m where m is arbitrary integer.

Homework Statement Let Q be the set of all rational numbers Prove bd(Q)=R Homework Equations The Attempt at a Solution Let x be a real number, then since the interval |x-r| contains both rationals and irrationals for arbitrary small r, so R is the boundary of Q. Is that right?

Arrrgh... Well, let's see... \cos\theta \sin\theta =\frac{1}{2}\sin2\theta =\frac{1}{2}\sqrt{1-\cos^2 2\theta} =\frac{1}{2}\sqrt{1-(2\cos^2\theta-1)^2} =\frac{1}{2}\sqrt{1-4\cos^4\theta+4\cos^2\theta-1} =\frac{1}{2}\sqrt{4(\cos^2\theta-\cos^4\theta)}...

You mean parameterize a line segment? Since the line goes through (-1,1) and (1,-1), then f(t)=(1+2t, 1-2t), so transform f(x,y) to f(t): (1+2t)(1-(1-2t))=2t(2t+1)=4t^2+2t=4(t+1/4)^2-1/4, so f(t) has min value -1/4 at t=-1/4, right? What about the other part...

How do you parameterize f(x,y)=x-xy? One thing I tried and was mentioned in my OP is polarize x,y, since x^2+y^2\leq1, therefore, let x=r\cos\theta, y=r\sin\theta, then 0<r\leq1, and from x+y\geq0, we can fix the domain for \theta as well. However, then...

Oops...Sorry for the double post. I was editing my original post but somehow it got reposted...Please ignore this one and use the other: https://www.physicsforums.com/showthread.php?t=177299 Thanks.

Homework Statement Find and classify the local and absolute extrema of the function f(x,y)=x-xy over the region D={(x,y)|x^2+y^2\leq1 and x+y\geq0} Homework Equations The Attempt at a Solution Critical points are where the first derivative (gradient) is 0. Grad(f)=(1-y, -x)=0 So critical...

Homework Statement Find and classify the local and absolute extrema of the function f(x,y)=x-xy over the region D={(x,y)|x^2+y^2\leq1 and x+y\geq0} Homework Equations The Attempt at a Solution Critical points are where the first derivative (gradient) is 0. \nablaf=(1-y, -x)=0...

There exists g(x)=m+b(x-a) such that f and g are equal to the first order, so you need to find such g (i.e., the value of m and b) according to the diff'bility of f.

Well, I understand the linear case but I can't quite figure out the bilinear case. Would you enlighten me on that? However, looking at another angle, wouldn't f(x,y) be in a general form of f(x,y)=Ax+By+C, where A is p*n matrix, B is p*m matrix and C is in R^p? Therefore, express P in terms...

Thanks for the reply. From what was given in the question, I don't see how to get a bound on |f(u,v)| where u, v are the unit vector for h and k respectively. First of all, I'm a bit confused by the notion of |(h,k)|. Let's say h is in R^2 and k is in R^3, then what does |((3,4)...

Homework Statement Let f: R^n X R^m --> R^p be a bilinear function. Prove that |f(h, k)|/|(h, k)| --> 0 as (h, k) --> 0 (zero vector in R^(n+m)). Homework Equations If f: R^n X R^m --> R^p is bilinear, then for x, x1, x2 in R^n, y, y1, y2 in R^m, a in R: a) f(ax, y) = f(x,ay) = af(x,y) b)...

Great! Thanks a lot man.

Oops...that was a typo. It's supposed to be ln(y)=3ln(x)+C. However, I haven't used the fact that the stream passes through point (1,1,5) yet...Can I just plug in the point into the equation: ln(1)=3ln(1)+C, then C=0. Therefore, is the equation of the stream ln(y)=3ln(x)? Thanks.