Recent content by miqbal

Are you familiar with Kirchhoff's Laws?

Sorry i thought it was an element of the symmetric group of degree 6.

If you don't wait you make an error like I did. When I applied the init conditions after solving for the complementary solution it caused the complementary solution to become zero.

The sigma format makes it much harder to see what's going on with the permutation. All it really is though is sending the nth element to the n - 3 element. So 4 goes to 1, 3 goes to 6, 2 goes to 5... \mu = \left( \begin{array}{cc} 1\ 2\ 3\ 4\ 5\ 6\\ 4\...

Yeah I did make that mistake. You can always check your answer by differentiating x and plugging it into the differential equation.

Not exactly. You did not use the binomial theorem correctly. The summation must start at 0. Also n = \infty.

Well my math up there is right too. In the special case where r = k (the number of trials). P(X < r) is P(we have r successes in less than r trials) = 0. So the math gives the same result of 1. Hope that helped!

Yeah i thought that made sense too but "doing the math" gave a different answer. Think about the same scenario again (a fair coin, look for 8 successes, k trials). Sum up the probabilities that we get 8 successes after k flips from k = 8 to infinity flips. Should this sum be 1? I have no idea...

Yeah sorry I was wasn't finished editing it. Its confusing no doubt though :smile:. I guess it's the complement that we will have less than r successes in k trials, but I'm not sure.

It's just summing up binomial random variables. We're summing over k - the number of trials. r is fixed so we are summing the probabilities that we have r successes for k = r,..., infinity trials. Hmm, let's take an analytic stab. Using the binomial theorem, (p + (1 - p))^{k} = \sum_{i=0}^k...

The Heini-Borel theorem only applies to euclidean spaces. Use the definition of compact.

Use the quadratic formula x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} No mistakes so far...

You get better at guessing particular solutions. First you have to get the complementary solution. Once you have that assume that the y is something like At2 + Bet. You're fine up until now as long as your complementary solution doesn't have the same term in it. Otherwise you will have to...

You need to evaluate \int R^{2}cos^{2}(\vartheta) \, d\vartheta Make sure you after you solve the integral you substitute back for the solution in terms of x.

You evaluated \int x \, dx Whereas your original integral was \int \sqrt {R^2 - x^2} dx