Recent content by hamsterman

The problem is with your second term, It should be 15/6 = 5/2, not 5/3. I didn't check if that solves the problem though.

I know it is true. That fact does not interest me. The proof using Hilbert system does.

Homework Statement Prove \neg x \vee x using Hilbert system. Homework Equations The logical axioms. I'm not sure if I should state them, or whether there is a standard set. It seems to me that different sets are used. Anyway, the ones with disjunction in them are: a \rightarrow a \vee b...

Not in general, I think. For example, \lim \limits_{x \rightarrow 0} \frac { \lim \limits_{y \rightarrow 0} y } {x} is undefined. If you picked y = x, you'd get 1. You could, however, pick y = 2x as that too satisfies the arrow. I'm not sure why it was done here. Anyway, e^x can be derived if...

Write down functions j(t) and z(t), then find the derivative of \sqrt{j^2(t) + z^2(t)}. Note that j and z are lines in R^2 space. jx is always 0 and zy is probably 2.

You'll have to be more specific. Do you know what a normal is? A scalar product? That <a, b> = 0 if a and b are perpendicular? That <a+b, c> = (ax+bx)*cx + (ay+by)*cy = ax*cx + ay*cy + bx*cx + by*cy = <a, c> + <b, c>?

In the line equation ax + by + c = 0, (a, b) is a normal of the line. Another way to write the same equation is <(a, b), (x, y)> = -c, where < , > is the scalar product. It follows from properties of < , > that if <p, q> = w then if you take any vector n, orthogonal to p, <p, q+n> is also = w...

You have a line Lp defined by, (a1x+b1y+c1) +p(a2x+b2y+c2)=0. Say that lines L1, L2 intersect at a point P, then P is both on L1 and on L2, thus a1*Px+b1*Py+c1 = 0 and a2*Px+b2*Py+c2 = 0. 0+p*0 = 0 so P is also on Lp. That's all there is to it. A family of lines means that Lp is not any...

The equations are correct. This is just a system of equations. You already have one variable expressed in therms of another. Now you only have to substitute that expression in the second equation and solve it. It will be a quadratic equation, when you clean it up.

It should, if it's an embedding from R, I think.

Well, since you have a curve, it is easy to write a function of distance from some point to (x0, y0). Then the least distance d(x, y) is the minimum of that function. Then show that d(x, y) = 0 means that y = x^2. Although proving that x -> (x, x^2) is an embedding is a lot more understandable...

In the example case it is easiest to say that there is a homomorphism from a curve to R, since being closed (or open) is a topological property. Another way should be to show that its complement is open by taking a point outside, finding its (least) distance to the set and then showing that if...

The book looks okay from what amazon allows to skim through and C is definitely a good language to know. Note though, that when you're learning C, you will need to know more about the internal details than you would in, say Java. It can be a pain, unless you enjoy it.

Here's what I'm thinking. Take a basis of vector space of square matrices V, such that each matrix has a single 1 and all other elements are 0. If you take a basis {v1, ... vn} and replace vk with c1v1 + ... + cnvn, you still get a basis, as long as ck is not 0. That is easy to prove, if...

Unless you excluded some condition, there is no relation between alpha and the other two. Any two right triangles could be composed like this.