Recent content by NatFex

Hi all, excuse the double post but since I just managed to find what I was looking for I thought I'd share in case anybody else finds seeing where the equations in the OP come from Quick scribble since I can't be bothered to make it as neat as my original diagram at the moment: The...

I was confused by your comment at first but you're right, I have it the wrong way around. OK, so the ##y##-axis (as per right-handed cartesian system) is going into the screen. The only way to make my diagram still relevant is to instead say that we're rotating clockwise around ##y##, so my...

Hello, I'm having a visualisation problem. I have a point in R3 that I want to rotate about the ##y##-axis anticlockwise (assuming a right-handed cartesian coordinate system.) I know that the change to the point's ##x## and ##z## coordinates can be described as follows: $$z =...

Indeed it doesn't! Formula yields 4 whereas there are 6 countable solutions. However, all I'm alerted to is the fact that I'm wrong and not how/why. I realize this is much to ask, so any input is appreciated.

Does it really matter if I arranged it as 5*4*3*2*1*5 or 5*5*4*3*2*1? (In the case of the latter, jobs 1 & 2 can be done by the same person, but given that multiplication is commutative and that, as I've said, the job numbering is arbitrary, does it make a difference?) I would like to see your...

Homework Statement [/B] The question is phrased in the following way: There are 6 jobs to be assigned to 5 people. Each job is assigned only to one person, and each person must have at least one job. How many different arrangements are there? Homework Equations In general, I would approach a...

Likewise, fixed my post a moment too late. Perfect, you confirmed what I thought was going on.

I saw your edit, now the confusion arises when taking ##f(2)##, which equals 1. I suspect my understanding of the significance of the values continuous distributions functions return is what's at stake here. EDIT: Of course, the fallacy is that with a continuous variable, ##P(X=x)## in exact...

But surely the probability that X takes a value of 1, ##f(1)## cannot equal 2? And yet I got the value of x=1 by integrating to ensure that the sum of probabilities never exceeds 1. That's where my confusion comes from.

I have a Stats exam on Wednesday and while I thought I was quite well-versed, I've gone back over to the very basics only to find myself confused at what should be introductory. Suppose I have a continuous random variable modeled by a probability density function: $$f(x)=2x$$ Obviously the...

Brilliant, just what I was looking for I completely forgot about looking at the complex numbers in rectangular form instead, cheers!

##z^{-1}##, but my proof in the OP only works for positive integers. I'm trying to use -1 as a base case to write a proof for negative numbers, but for some reason I cannot work out how to apply a similar strategy (expand then use compound identities) It doesn't matter if θ is positive or...

Yeah, but I can't prove that that last step (raising the complex number to a power of -1) will make the angle/argument inside ##\cis## negative I know that ##\frac{1}{r\cis\theta} = (r\cis\theta)^{-1}##, but -1 is outside the boundaries for n that I have proved in my first post :( I know it...

Or basically anything that isn't a positive integer. So I can prove quite easily by induction that for any integer n>0, De Moivre's Theorem (below) holds. If ##\DeclareMathOperator\cis{cis} z = r\cis\theta, z^n= r^n\cis(n\theta)## My proof below: However I struggle to do this with...

This has been asked already, but it was either answered incorrectly or I'm just not understanding it right. I was under the impression that albedo is the ratio of radiation reflected off a body compared to the total incident on that body. Hence black bodies have an albedo of 0 and more...