Recent content by deiki

You can use it to prove the Poincaré Lemma and compute the local potential of a closed differential smooth form. You can use it when you are computing the derivative of an integral on a manifold that depends on a the parameter you're conducting a derivative with, like the general case shown in...

Don't go with i = √(-1). It gives you awful stuff like the following : i = √(-1) = √(-1/1) = √(1/-1) = √(1)/√(-1) = 1/i ... ouch. i i = -1 works better as a starting point. Baby Rudin starts this way : Let a, b, c, d be real numbers. By the definition, x = (a,b) and y = (c,d) are...

I'm not sure the proof I'm giving here is relevant to the topic, but since there are tons of definition of the trigonometric functions with different proofs to find their properties ( which end up being the same at the end ), you might want to have a look at this one. This one might help if you...

uhm, that's just Taylor's theorem. The proof is actually on wikipedia.

My explanation is quite informal, but I hope it helps. Well it's not a constant, but roughly speaking, it's just a positive number when you approach given conditions ( usually the Weierstrass limit conditions ) that will grow bigger and bigger ... and bigger ... and bigger ... and bigger ...

just use (n+1/2)! = Γ(n+1/2 + 1 )

using series from the general case actually works. to plug the 1/2 into this equation, you will need to study a special case of the gamma function for the factorial. Once you do this, the whole 2^(m+2n)∙n!∙(n+m)! should simplify nicely for m=1/2...

Let A be a ( large ) positive real number The transform you specified will lead you to integrate exp( - i * w * t ) with t ranging from -A to A, and letting A approaching +∞. When A approaches +∞, this will give you a function similar to one of these ...

Yes, you can expand the exp function into powerseries. Plugging the properties a*+b*=(a+b)* and a*∙b*=(a∙b)* into these powerseries will lead you pretty quickly to [exp(z)]* = exp(z*) This work for any holomorphic function you can expand into powerseries with real coefficients.

Hölder inequality.

http://a.imagehost.org/0247/chern.jpg :biggrin: :biggrin: :biggrin:

Geometric intuition can lead you to algebraic conclusions as well as algebraic calculations can lead you to geometric conclusions. The problem is that many geometric results are just special cases of algebraic computations, and you will often need algebraic results which can't be described...

try : solve(sign(factor(x^2-1))=-1,x)

( I assume a,b ≠ 0 ) | x - a | < ε' hence | x∙y - a∙y | < ε' ∙ | y | | y - b | < ε' hence | y∙a - b∙a | < ε' ∙ | a | you sum both inequalities | x∙y - a∙y | + | y∙a - b∙a | < ε' ∙ | y | + ε' ∙ | a | you use the triangle inequality | x∙y - a∙y | + | y∙a - b∙a | > | x∙y - a∙y + y∙a -...

you can define the gradient operator such that : d \Phi = \left< grad \Phi , d\vec{r}\right > knowing that in spherical coordinates : d\vec{r}\right = \vec{e}_{r} dr + \vec{e}_{\theta} \cdot r d\theta + \vec{e}_{\phi} \cdot r \cdot sin(\theta) d\phi then you should find what you want.