@soupless Thanks. I think someone else is reporting these as well to the mods, because I'm seeing recurring versions, but each time the account is from 2014, the questions answered on MSE keep differing. I don't know how this happens!
Hello @Koro! I haven't come here for some time, so it makes sense to look at things now.
:SarveshRavichandranIyer The axiom of completeness answers my question. But now, my question is: is the axiom of completeness enough or do we need other axioms/properties aside from this?
@S.M.T In response to this query, which is coming quite late : If the general case gives the answer 0, then naturally every special case will also have the same answer. The answer to the question which you posted and involved vectors, is general : it doesn't assume any position for O,P but only what's given. How do you justify that you are tackling the "general" case? That might be a question I cannot answer very nicely, unfortunately.
@soupless I believe completeness is very fundamental. I don't know how much, but I remember reading somewhere that completeness goes down as a category-level axiom : a very deep one. That usually means that anything also doing the job is either implying completeness very superficially or a very strong condition.
Completeness alone is enough, I think.
I also think that one can't do without it as well, but there I could be wrong. I think TSF might be able to say something about this.
@soupless It has links, but I'm referring to category-theory. There is some operation at that level called functoriality, and the "completion" is an functor at that level if I am not wrong.
@Koro I saw your answer, it's correct. To make a further comment, I believe I've seen the same argument in the proof of Ostrowski's theorem : which shows that the only non-trivial absolute values on Q are the usual one and the p-adic one.
Unfortunately, I can't see the question. The title seems to be If $p$ is an odd prime, $n \ge 1$ is a natural number, then and I can't get the next part
I need some hints for this problem from Dummit and Foote. (edit: added the full question verbatim for context)
Let $p$ be an odd prime and let $n$ be a positive integer. Use the binomial theorem to show that $(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n}$ but $(1+p)^{p^{n-2}} \not\equiv 1\pmod{p^n}$. Ded...
@soupless I looked at that question. What you did makes sense because you under-approximated the sum by the integral, and the integral was close enough to the sum. To understand or bound the difference between the integral and the sum, you'll have to use the mean value theorem, or use error formulas from numerical integration. I seem to remember this approximation as a "rule" in my time, and I've forgotten what the error formula looks like.
If you can't see a question, you can click the "Try a Google Search" and there should be a result. If there was a result, find the options (three dots beside the clickable link) and use cached
This page contains some details around how much the error is. According to the formula, your error is at most $M \frac{(b-a)^2}{n}$ where $n$ is the number of divisions, $a,b$ are the integral limits and $M$ is the maximum of $|f'|$ on $[a,b]$.
Now we have to calculate these quantities from the question.
So $a=1,b=2006, n = 2006$, and $M=0.007$ : that comes to around $15$ so @soupless the error of the approximation is bounded by $15$. It is much better than that because the derivative actually decays rapidly at $1$ before stabilizing, so $|f'|$, which is very high at $1$, plays a problematic role.
@SarveshRavichandranIyer: I have one idea. We consider $|x|$ on [-1,1] and then try to construct a Weierstrass like function, which is known to be differentiable nowhere.
But that will not work yet. Because we have so many 'pointy' ends.
@soupless Yes, that comes to around 1.901, as expected. So this means that the bound is even worse than I initially anticipated.
I got the wrong bound of 0.007 : with that, the answer was around 15. Now it will be much larger, so the integral approximation is likely very good only because the derivative doesn't fluctuate heavily i.e. $f''$ takes very small values.
But this isn't captured by the formula we have, and one might need a Taylor to improve it.
Wait, the function I used to approximate the sum was $$\frac{2008^{2x/2007}}{2008^{2x/2007} + 2008}.$$ Maybe this can improve it, or is it the one we are already using?
Oh, I forgot. The original function was $$f(x) = \frac{2008^{2x}}{2008 + 2008^{2x}}$$ so the result will be changed.
Ah good, that's the one I was working with! So I hadn't made a mistake, and the error is, in fact, something like 15.346. If you take the original function, you will get this bound.
The second function you mention, is the one being integrated, right? So it seems that this will capture the error.
I think it is, since we can use the chain rule where $u = 2x/2007$ and the derivative of the first function is equal to the derivative of the second function, scaled by $2/2007$.
It can still be used for the second derivative, so yes. It is.
@Peter Interesting, do keep up with the search. I wish I could participate. I'll let you know when I'm free.
@soupless Oh, ok, I'll verify that. I did this by hand so I'm sure I got something wrong along the line.
@soupless Ah right, I'd made a mistake. On wolfram, the max is 0.00189, indeed. So combining that with the other terms, gives 3.78
or around that.
This seems correct to me.
@soupless I'll have to leave, unfortunately. It is quite late in the night so I'll wrap up and come back tomorrow when I have time to take care of the situation better.
Solving for the number of divisions to get the integral accurate up to 0 decimal places, if that makes sense, is 2003 divisions.
What I did was use the $2x/2007$ function, solve for the number of terms where $$\frac{b - a}{n}\Big(f(b) - f(a)\Big) \leq 1$$ and letting $a = 1$ and $b = 2006$ and solving for $n$, we have $n \geq 2002.988806$ which is almost $n \geq 2003$. To be safe, maybe $n \geq 2002$. 2006 is close to 2002, so it's okay to round off?
