the tree

Your method can't tell you? Mine says that it's just an unfinished pyramid, not representative of an actual structure - the 13 steps to symbolise the original 13 states and the eye of providence to go with the 'one nation under god' motto. Although I'm assuming you mean 'architecture', since there aren't any architects on the $1 bill.

My method suggests three answers. Google, IMDB and Wikipedia. That's right! Three answers! Beat that.

See, the thing with your 'method' is that it doesn't really exist. What you have is a picture of 228 dots, and some lines in between them. I could just as well say that my 'method' is urm... the Julia Set and a picture of Kaya Scodelario. Far superior because it does just as much as your method - and faster, and has a picture of Kaya Scodelario.

I've been thinking about the original question, there was a lot of technical language in your proposed answer and TBH I don't even understand what a sheaf is. To me, you start doing geometry once you involve a meaningful notion of distance: for it to be meaningful there should be a distance function where distances between distinct pairs can be equal but aren't always - that way you can have shapes and congruences that you'd expect from something called geometry. Of course that may be a really simplistic definition but it covers all the things I've known to be called geometry.

It's definitely worth understanding that "gravity affects light" does not imply "light affects gravity". Overcoming gravity is very different to actually weakening the gravity itself and there is no evidence that you've done either.

But don't be fooled into thinking that everything you see on arXiv is any good.

I doubt the OP meant it literally.

I don't really see the issue here. Use Gaussian elimination and then back substitution as per normal.

Academic papers tend have well, data, in them. Ideally that data should be comprehensive, statistically significant and above all, replicable.

Well, you can squeeze around that "I speculate that Humans know nothing" or "I, speaking as a Martian..." &c.

The sequence is strictly increasing since [imath]\forall n \in \mathbb{N} : (n+1)^2 = n^2 + 2n + 1 > n^2 [/imath]. The sequence is not bounded above since [imath]\forall b \in \mathbb{R} , n>b : n^2 > b^2 \geq b[/imath]. And it's well known that a strictly increasing sequence is only convergent if it is bounded above. //

A sequence [imath]\{ x_n \}[/imath] is said to converge to a limit [imath]l[/imath] iff for a given [imath]\epsilon>0[/imath], [imath]\exists N=N(\epsilon )[/imath] such that [imath]\forall n > N \; |x_n - l | < \epsilon[/imath]. Now, this is very important so read carefully: A sequence diverges if and only if it does not converge. Got that? So you need to look at the negate of the definition of convergence. To be really, horribly slow about this, the negate of: "for a given [imath]\epsilon>0[/imath], [imath]\exists N=N(\epsilon )[/imath] such that [imath]\forall n > N \; |x_n - l | < \epsilon[/imath]." is not and never will be: "for a given [imath]\epsilon>0[/imath], [imath]\exists N=N(\epsilon )[/imath] such that [imath]\forall n > N \; |x_n - l | > \epsilon[/imath]."

Presumably we'd be gaining that atmosphere back somehow, if it were only being topped up by volcanic activity then the net mass of the Earth would be slowly going down, which I'm sure we'd have heard about.

Well no, it isn't. Look back at your definitions of convergence and divergence.

You only really need to pick a big-N if you're proving that something converges. For this I'd recomend finding a subsequence that is known to diverge and assert the lemma that a sequence with a divergent subsequence must be divergent itself.

One thing that DeCartes got right (amongst many things that he got so very wrong) was cognito ergo sum: so long as we can consider the possibility we can be sure that somehow, somewhere, something is going on - so we have a definite starting point in reality. Then we have a priori and axiomatic reasoning that we can build a sort of knowledge from (albeit vacuous), so we know something. And if everything we believe about the world turns out to be wrong then that's not a massive problem, we'll just scrap what we've got and start again - we've done that before and called it the enlightenment but we can always do with another one.

Okay, first you need to need to chill with the ellipses, you're using them for the wrong reason and they are making your writing incredibly hard to read. Then try to focus on one thing at a time. It's true that the sum of human knowledge is barely a drop in ocean of potential knowledge and even then it is one polluted with uncertainty and formed out of clouded interpretations - but scientists and philosophers and statisticians and engineers and historians are all working on it. Now, to some scientists a rock may just be a rock - though to a geologist or a palaeontologist it may be the closest thing they have to a record of millions of years worth of exciting history. Scientists aren't all that likely to see something and say it's just what it is, especially when they have the tools to learn something from it. Now as it happens, if you are willing to take the leap of faith to accept inductive empiricism and all that - then it turns out that we do know an awful lot and the fact that it pales to insignificance when pitched against the whole universe should not be discouraging but rather a reason to go on learning with all the more enthusiasm.

ε is kind of chosen for you, it should really be the first line of your proof. You're along the right lines you just need to put your thoughts in order.

So I did. Fixed and now the answer makes a lot more sense, thanks.

First up, partial fractions to make this vaguely approachable. [math]y=\frac{k x}{1+x}=k-\frac{k}{1-x}=k\left( 1-\frac{1}{1-x}\right)[/math] Then some substitution: take [imath]u:=\ln(y)[/imath] and [imath]v:=\ln(x)[/imath]. [math]e^{u} = k\left( 1-\frac{1}{1-e^v}\right)[/math] Take logs. [math]u = \ln(k) + \ln(1 - \tfrac{1}{1-e^v} )[/math] Finally, differentiate. [math]\frac{du}{dv} = \frac{ \tfrac{d}{dv} \left( 1-\tfrac{1}{1-e^v} \right)}{1-\tfrac{1}{1-e^v}}=- \frac{ \tfrac{e^v}{(1-e^v)^2} }{1-\tfrac{1}{1-e^v}}[/math] Throw [imath]x[/imath] back in and simplify as much as you feel like... [math]\frac{du}{dv} = \frac{-x}{(1-x)^2 (1-\tfrac{1}{1-x})}= \frac{-x}{x^2 - x}= \frac{-1}{x-1}[/math] And voilà! [math]\frac{d\ln(y)}{d\ln(x)}= \frac{-1}{x-1}[/math] And no, I have no idea why you would want to do that, or why it would be relevant to anything.

It's kind of beyond me how this is relevant - it's an actual theory, doesn't matter where it originally came from.

Severian- If I may ask, how did you become a reviewer? Is it a full time thing or are you doing your own stuff as well? Is it a fun job?

What do you mean by Carpenter's Logic? I think this is something I don't know about.

ffs

Well if we were headed in the right direction then we'd get there eventually, but it'd take so long that it'd feel a little pointless. Still don't think the stopping should be an issue. If we could get say, the moon to undergo nuclear fusion then we should have enough energy to reach escape velocity.