Recent content by jbstemp

I'll take a go at 8b. Let ##p(t) = T^3-2##. Notice that ##p## has not roots over ##\mathbb{F}_7## which can be seen by simply evaluating ##p## at each element of ##\mathbb{F}_7##. We claim then ##p## is irreducible. If not then ##p## can be factored as ##p = f g## where ##f##, ##g## are...

Problem 2(b)

I'm so confused by this post. Are you talking about how ##e^{i\theta}## is a circle in the complex plane with radius ##1##, or how the series expansion for ##(1+\frac{1}{n})^n## is ##e-\frac{e}{2n}+O(\frac{1}{n^2})##, or something else? Regardless, e certainly is an amazing number and pops up...

I don't think there is a way to count the number of spanning trees of a graph if you only have the number of nodes and branches. However, If you are able to label the vertices and edges of a graph then there is the "deletion-contraction" theorem which says that the number of spanning trees of a...

So you have (x^2+y^2+y)^2 = x^2+y^2. After converting to cylindrical coordinate and a bit of rearranging you have r^2 + r(sin\theta-1) = 0 Solving for r using the quadratic formula will give you r = 0 and r = 1-sinθ.

You need to take into account that r depends on theta, and as such you can't simply integrate it from 0 to 2 like you would with a cylinder of constant radius. After a bit of algebra you should find that r is bounded between 0 and 1-sinθ, so the integral you want to evaluate is: \int_0^{2\pi}...

The series is divergent.

The displacement of the center of mass of an object should be the same as the displacement of any other point, assuming the object is rigid (doesn't stretch or contract).

I think technically you could evaluate those types of integrals, but you won't really get any thing with meaningful results.

Everything you posted seems to be correct. Try typing this into the graphing software, the points should match up. f(x) = 3cos(2x-\pi/2)+1

I really don't understand this question. Could you rephrase it?

The 3rd order taylor series for sin(x) would only be up to the x^3 term. In the remainder formula, n refers to the number of terms, not necessarily the powers. In this case x is the first term, -x^3/3! is the 2nd term, and x^5/5! is the 3rd term.

It's standard to evaluate indefinite integrals as the anti derivative of the function plus a constant of integration. So I think ∫sin(x) dx-∫sinx(x) dx would just be equal to an arbitrary constant (as you said).

Hey everyone, I'm currently a first year undergrad physics major with a minor in math I have a wide range of interests but am particularly fascinated by dark energy and cosmic inflation (who isn't). Anyway, I look forward to my time here and hope I can contribute and learn from these forums.