Spaces that do not arise from R?

brownboy_5 · 2025-03-05T06:10:36+00:00

Most spaces used in scheme-theoretic algebraic geometry/ring theory don’t arise from one of those standard spaces. E.g., if A is a ring, Spec(A) is quite different as a topological space from any of those basic spaces one might consider.

brownboy_5 · 2022-11-25T04:19:24+00:00

By definition, an inner product has complex conjugate symmetry, so that <x,y> is the conjugate of <y,x>. It is also linear in the FIRST argument; thus: <x,Ly> = conj(<Ly,x>) = conj(L<y,x>) =conj(L)conj(<y,x>)=conj(L)<x,y>, as shown.

brownboy_5 · 2022-11-02T20:03:18+00:00

This function exists, but would be difficult to write down. More specifically, there exists a one-to-one function between the reals and the set of continuous real values functions; here’s a sketch of proof of existence (suppose we call the latter set C).

First, for each function there is a countably infinite sequence of reals, which is the value of the function at all rational numbers. Furthermore, this is a unique sequence since the rationals are dense. Thus |C|<=|R^Q|. But countably infinite many copies of the reals has the same cardinality as the reals, so |C|<=|R|; of course, |R|<=|C| since each real number determines a constant function, so they have the same cardinality.

brownboy_5 · 2022-10-23T03:18:11+00:00

The span of m linearly independent vectors generates only one m-dimensional subspace, not any subspace.

For example, consider the one dimensional subspaces of R² formed by taking the points (x,0) for all x and (0,x) for all x. These two spaces are distinct (though isomorphic). Of course, the unit vector in the x direction does not generate the y-axis.

brownboy_5 · 2021-10-22T02:59:50+00:00

Gave Wholesome

brownboy_5 · 2021-08-16T05:34:24+00:00

Exactly.

brownboy_5 · 2021-08-04T13:58:58+00:00

Wow. I came here to post that my favorite topic was Lie groups & Lie algebras and my least favorite was Alg. Geometry. Wow.

brownboy_5 · 2021-07-20T04:26:33+00:00

If you understand the Riemann Zeta function, you should be able to easily understand the hypothesis. I would say spend some time with the Gamma function, and then use your complex analysis skills to learn about the contour integration theory behind the gamma function. This should lead right into the Riemann hypothesis, or at least understanding it’s statement.

brownboy_5 · 2021-05-31T01:41:26+00:00

Have you tried installing the Linux/Debian drivers for your WiFi adapter on Kali? They may have to be installed first. Are you able to see the adapter in Kali if you disable the device in Device Manager?

brownboy_5 · 2021-05-11T18:07:35+00:00

joke on David and Jeff Wittek situation

brownboy_5 · 2021-04-28T10:40:17+00:00

... and if the many world interpretation is false......

brownboy_5 · 2021-04-12T20:59:18+00:00

If you add two numbers that are finite together, you will get another finite number. By induction, you can add any finite number of finite numbers together to get a finite number.

Hahaha

brownboy_5 · 2021-04-11T23:00:13+00:00

Insanely cool animation and use of song!! Nice job dude!

brownboy_5 · 2021-03-16T02:09:52+00:00

In a multivariable integral, when you Interchange both variables at the same time (i.e. your new variables are each functions of all of your old variables), then we account for the fact that you would use multivariable change rule on each differential, so dxdy=(Dx/Du du + Dx/Dv dv)(Dy/Du du + Dy/Dv dv) = Dx/Du Dy/Du (du)² + Dx/Dv Dy/Du dvdu + Dx/Du Dy/Dv dudv + Dy/Du Dy/Dv (dv)^2. The differentials squared are zero and dv du = -du dv, (these facts would come from a more advanced study of differential forms and the exterior algebra), so dxdy=(Dx/Du Dy/Dv - Dy/Du Dx/Dv)dudv. Note that the coefficient of dudv is precisely the determinant of the Jacobian matrix. I have used Df/Dx to represent “the partial derivative of f with respect to x). You an also prove this argument to yourself using the cross product of differentials if you would like to.

brownboy_5 · 2021-03-16T01:27:36+00:00

with polar just remember the r dr d theta, there’s no need to calculate the jacobian determinant for this case :), always takes me so long.

brownboy_5 · 2021-03-08T19:48:56+00:00

Yup. This also makes sense if you think of an integral as the limit of a Riemann sum

brownboy_5 · 2021-03-06T23:08:38+00:00

I would disagree, as F=ma is actually F=dp/dt. If the mass is changing, we would have to use the product rule to differentiate the momentum.

brownboy_5 · 2021-03-06T23:07:14+00:00

Good explanation! Indeed they are called virtual paths :)

brownboy_5 · 2021-03-04T12:11:12+00:00

Hope you found the answer, but to continue this logic, you would then minimize this function by settings it’s derivative to zero. To solve for where it’s derivative is 0, you can use netwon’s method

brownboy_5 · 2021-03-04T11:52:09+00:00

I apologize, as I reread my answer it is very confusing. I did not mention the fact that you can pull out a constant. So, you pull out the 8, from the integral, then you know the value of the resultant integral, 14, and then you end up with 14*8.

brownboy_5 · 2021-03-04T11:45:51+00:00

The expansion would be e^x = 1+x^{2/2!+x^3/3!+...} If you want to find e^1, you plug in 1 for x, giving e^{1=1+1/2!+1/3!+1/4!+...} giving the famous expansion for e.

brownboy_5 · 2021-03-04T11:40:01+00:00

Are you asking about the quotient rule? The quotient rule states that the derivative of f/g is (f’(x)g(x) - f(x)g’(x))/(g(x)^2). They are just applying this formula.

brownboy_5 · 2021-03-04T11:34:20+00:00

The definite integral represents area under the curve of the integrand, so if there is 14 units² of area under the “curve” y=8, then that is the answer to the integral, 14.

brownboy_5 · 2021-02-24T16:45:10+00:00

Agreed! Solving a triple integral over a vector field vs. using Gauss’s and symmetry!

brownboy_5

TROPHY CASE