Recent content by Dr Avalanchez

I'm still not sure what you mean, but these are some criteria I know of: * the eigenvectors are a basis for the vectorspace * all the algebraic multplicities equal the geometrical multiplicities (the vectorspace is a direct sum of its eigenspaces) * the characteristic polynom has no multiple...

Try with induction by blocks (blocks of 2^k).

It only means that your function is solvable (and unique) in an open interval round x. It does not tell what or how you should solve it. It only tells that your quest for such a function will not be fruitless. In most cases, it is sufficient just to know that such a function exists, without...

You should really try to figure this out yourself. What's the difference between the two equations?

Implicit function theorem?

Aaargh... nearly there... y^2=-x^2+C Does this ring a bell?

If I understand your problem correctly, this is just an application of the implicit function theorem? f(x,n) is your function, P=(a,b) is a point satisfying f(P)=0. If df/dx<>0 in P, then there exists a unique function g so that f(g(n),n)=f(x,n) (so g(n)=x) in an open interval round x.

sin(2x)=...? cos(2x)=...?

Just apply Simpson's formulas.

How to start this: sin(3x) = sin(2x+x)

a function is even iff f(-x) = f(x), for example cos(x) a function is odd iff f(-x) = -f(x), for example sin(x) exp(-x) <> exp(x) <> -exp(x), so exp(x) is neither exp(-x)+exp(x) = exp(x)+exp(-x), so exp(x)+exp(-x) is even (=2cos(x)) exp(-x)-exp(x) = -(exp(x)-exp(-x)), so exp(x)-exp(-x) is...

No, I was just pointing out that you can have all sorts of norms, so it should be stated what vector space we're talking about, and which norm. It was a critique on Matt that mathematicians do make a distinction (unless it's clear what we're talking about, like here).

I do not agree on that. ||.|| is a norm and | . | is the absolute value. norm() applies to vectors, abs() applies to scalars. abs() is a well defined function for real numbers (and the complex analogon), whereas norm() is not. A function is a norm in some sort of vector space iff it satisfies...

You could explain it pure algebraically (this holds for higher dimensions): Starting by the proof of the Cauchy-Schwarz inequality we have: \left| {x \cdot y} \right| \leqslant \left\| x \right\|\left\| y \right\|. This is of course the same as: -\left\| x \right\|\left\| y \right\|...

Probably, but I'll have to think about it. (don't hold your breath, I'm in the middle of exams)