Recent content by hypermorphism

Mathematica (pricy) or Maxima (freeware).

These are the triangular numbers. The nth triangular number is given by the formula n(n+1)/2. For more information, see http://mathworld.wolfram.com/TriangularNumber.html and http://en.wikipedia.org/wiki/Triangular_number .

It may just be experience. Like the way most of us can do integrals, without having to go through pen/paper or even acknowledge intermediary steps.

If by R^2, you mean a 2-dimensional vector space over R, and by R^3, you mean a 3-dimensional vector space over R, then the correct statement is that R^2 is isomorphic to a subspace of R^3 (one possible isomorphism is mapping all pairs (x, y) to 3-tuples (x, y, 0)), but R^2 is obviously not...

Use the dot product.

One differentiates functions, not numbers. Do you mean you want to know the derivative of the identity function? Over what domain? Or do you want to know the derivative of the function that always returns 1?

If you are attaching the vector to a point, you actually need to find two separate results: the new point the vector will be attached to after rotation, and the new orientation of the vector. If we call the position vector of the point the vector is attached to p, and the vector v, then these...

Sometimes the second derivative yields no information about concavity. Consider the two different functions f(x) = x^4 and g(x) = -x^4. These two functions have different concavities, but if evaluated at (0, 0), their second derivative is the same. A higher derivative will reveal their different...

I have never found anything like Spivak (although Apostol and Courant are great for calculus as well). "Calculus on Manifolds" is pretty much the definitive treatment of vector calculus for those planning to study differential geometry. As said above, his comprehensive treatment of differential...

No and Yes 1. It is not always the case that one can find a closed-form expression for the inverse function when an inverse exists. For a simple example, take the function f(x) = x + cos(x). It is obvious that the function is everywhere 1-1 (we pretty much label each cosine with its argument to...

The RHS is factorable into a function of x and a function of y.

I think that one's also provable through combining Brouwer fixed point with Borsuk-Ulam. Yep. It seems so simple, but it seems to elude capture unless one calls on sledgehammers. :confused:

If we are working in Euclidean space, the theorem is just Invariance of Domain.

Mathwonk described the theorem in his post. For polynomials of the second degree, the result is trivial, so it is generally called upon to reduce higher degree polynomials with rational roots. Plug in rational combinations of factors of the constant term over factors of the leading coefficient...

The method of undetermined coefficients relies on all linear combinations of the linearly independent derivatives of the RHS. You really don't know yet whether there are lower power terms on the LHS that have simply canceled out. So your Y should be Y(t) = At^2 + 2Bt + 2C = At^2 + Dt + E. Plug...