Recent content by Office_Shredder

@elias001 I'm sorry your post is long and I'm not sure what all your questions are but I'm going to start with yes, you correctly understand the diagonal property and for assumed exercise 1 (which you it in bullet 3). This isn't asking you anything fancy. Fix n. Let ##x_{\infty} =(1/k)## be the...

You want to construct a bunch of convergent sequences that all converge to the same value such that any diagonal you try to pull out is guaranteed to not converge under the definition of convergence. Here's an example. Suppose we look at functions ##\mathbb{R}\to \mathbb{R}## and define...

You very rarely want to try to show you violate an axiom directly. You already know that if a metric space can give the convergence definition then it must have the diagonal property mentioned. It is fairly easy to just construct a bunch of converging sequences which don't satisfy the diagonal...

To do the diagonalization argument the goal is to construct sequences that all converge to the same point (let's make it ##(0,0,....,0,...)## for simplicity) but you cannot pick one entry from each sequence and have this new sequence converge to 0. I would recommend looking at very simple...

Either you have some theorem that just surprisingly crushes this, or my guess is you're going to have to think about how you can approximate irrational numbers with rational numbers and use what happens to this with rational numbers. So the first thing I would do is think about what happens if...

You should think of a single order as the order happening and then a bunch of days until the next order. So the first data point you have is you spent 158 dollars and then waited 20 days until the next order. Then you spent 158 dollars and waited 15 days until the next order. The very last...

C12 should be something line sum(C2:C8)/(D11+DATEDIF(A2:A8))/30) Adding D11 to the DATEDIF (before converting into months)

Your average is too high because you're taking two orders as your endpoints. Here's a conceptual example: suppose you have two orders that are actually one month apart, 30 days, for 250 dollars each and you add up the values in rows 2 and 3 just like you did here. Then you get...

Well hold on, even the distribution is sketchy if you don't have absolute convergence. I would be pretty nervous distributing $$\sum \frac{(-1)^n}{\sqrt{n}} \sum \frac{(-1)^n}{\sqrt{n}}$$

I'm just summarizing what has been said but maybe rewording it helps. I'm going to just use power series around 0 for notational simplicity. Suppose ##\sum a_n R^n## converges for some radius ##R##. Then ##|a_n R^n|## must converge to 0 and hence it's eventually bounded by 1. Then if n is...

It's trivial that you can get two points to be linearly dependent. For example the square with vertices (0,0), (0,1), (1,0) and (1,1). (0,0) Is linearly dependent with every other vertex.

Bullet point 2 mentions f being in ##C^2## which means its second derivative is continuous. It's unclear to me if that's an assumption for this entire section or not. It might help to post the entirety of what is being done so we know what assumptions you're starting with.

It seems exceedingly unlikely because most of the techniques Wiles used were from branches of math invented well after Fermat died.

A far stupider example is to just take a line and draw a squiggle that goes up and down at a diagonal. 1=sqrt(2) can get you to a lot of contradictions really quick

It sounds right to me but I think there are simpler ways of describing it. Don't use the word pivotal column since the theorem doesn't call them that: If rank(B)=rank(B') then B' contains the all the columns mentioned in point (b) of the theorem for B, and then (d) uniquely describes how b...