Hey everyone. Should I generally call the popular Banach-Tarski statement Banach-Tarski theorem or Banach-Tarski paradox? There are generally more mathematicians agreeing with the truth of the axiom of choice, but does that really mean it'd be better to call it a theorem?
@robjohn I'm really grateful that SE has users like you too, because - Let it be noted - 60,317 rep doesn't make anyone famous shit or smart ass.Thanks for answering the question.
So, there does exist proof that it can't be solved in poly-time. b'coz gerard t'hooft recently said that quantum mechanics has an underlying deterministic structure. In case that's true, larger quantum computers under construction won't work
It is very sad that Rudin's Functional Analysis hardback is out of print. There are still copies being sold though, and the paperback is still in print.
How i can count percentage of my win rate for my games? win games / (wins+loses) * 100 ? so why then if i have 1 win and 100 loses its show 0.99% why not 1%?
@BalarkaSen Yeah, and I worked with another mathematician on that article. Believe me, it's crazy awesome, not because it's mine. If you see it you'll wonder if it comes directly from Ramanujan.
@robjohn Okay thanks, one thing. Also is it equivalent to state that a sequence has no supremum and a sequence is not bounded, using the completeness axiom?
Differentiation techniques - take $$\frac{(x-1)^3}{2x-1}$$. I started by multiplying out the numerator and got into a mess. Clearly chain rule seems better for this. Is it fair to say that chain rule is generally preferable to multiplying out powers?
When the derivative of a function is a rational and one wants to find the second derivative in order to ascertain the nature of the extrema, is it really okay just to check against the derivative of the numerator of the first derivative?
That's what I thought. I now see why it is relevant in my particular case though. Because the denominator of the first derivative is $(2x-1)^2$, so it's always positive. Smart move!
@topper I knew you were on the right track when you factored the numerator. You could use L'Hospital on the fraction (leaving the $e^{2x}$ out) or factor the denominator
@topper: Just factor $a = a_1 \cdot a_2$ and then try to write the quadratic polynomial as $(a_1 x + \dots) \cdot (a_2 x + \dots)$. Since there are usually several ways to factor $a$, you'll have to try out and consider the other factors ($b$ and $c$) too when choosing $a_1, a_2$.
I'm not looking for a solution please, but for $$\lim_{ x \to \infty}\frac{e^{2x}}{2x-1}$$ (same thing but tending to infinity), would you invoke L'hopital as 1) it's indeterminate $\frac\infty\infty$ and 2) I can't do the "take the highest powers of x and compare coefficients" as the numerator is "problematic". Just looking for confirmation of my strategy, not a solution please :)
When I cancel part of the denominator as per above, is it fair to say that I should use the original denominator, not the "new" denominator when deciding what the domain of the function is i.e. which values of x would give zero in the denominator?
Following on from the above function, let's say it's undefined at $x = 1$. Then I go to find the limit of the function as x tends to 1, to understand if I have an asymptote or a discontinuity. When I do my substitution of 1, can I then use the function that I've cancelled, or do I again have to use the original function?
@JasperLoy I was hoping there'd be a bare bones one (i.e. just put equations inside $...$, use ^ for exponentation etc.) tailored for newer users of this site
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Let's take $$f(x) = \frac{2x^2+x-1}{x^2-1}$$ which becomes $$ \frac{2x-1}{x-1}$$ after simplification. I know I can't use the simplified form when checking discontinuities - is there any other area where I should be using the original form?
Hi @DanielFischer could I ask one thing. If you have an increasing sequence $x_n$ which is bounded then by the monotone convergence theorem it follows that the sequence converges to $\sup_n x_n$, is it true then that max x_n = sup_n x_n?
@topper There's no need to look at it before simplification, but if we insert $x = -1+u$, we have $$\frac{2(u-1)^2 + (u-1) - 1}{(u-1)^2-1} = \frac{2u^2-3u}{u^2-2u},$$ and the limit of that as $u\to 0$ is $\frac{-3}{-2} = \frac{3}{2}$.
@DanielFischer Awesome! I think I got a bit thrown when I realized that I couldn't use the simplified form to see were the function has discontinuities. Suddenly I was like, "when can and can't I use the simplified form?"...
@topper You can use the simplified form to see true discontinuities. Only for artificial pseudo-problems you need to see where the original denominator has zeros, but the simplified function is continuous.
@JohnJack Sure. There are (positive) integers arbitrarily close to integer multiples of $\pi$, so for example $0$ is an accumulation point.
@JohnJack No, $\tan x = 0 \iff x \in \pi\cdot\mathbb{Z}$, so for an integer $n$, $\tan n$ is zero only for $n = 0$. But $\tan n$ comes arbitrarily close to $0$ for integers $n\neq 0$.
$\tan$ is continuous on $\left[ \left(k-\frac{1}{2}\right)\pi,\left(k+\frac{1}{2}\right)\pi\right]$. The continuity of $\tan$ at $k\pi$ plays a role when we deduce that $0$ is an accumulation point of $\tan n$ from the fact that there are multiples of $\pi$ arbitrarily close to integers. To just deduce that $\tan n$ has accumulation points (in $\mathbb{R}$, the existence of accumulation points in $\overline{R} = [-\infty,+\infty]$ is clear), weaker assumptions suffice, but the continuity and
@DanielFischer Let $a$ be a positive real number. Define the sequence $\{a_n\}$ by $a_1 = a$ and $a_{n+1} = \frac{1}{a_n} + a_n$ for $n \geq 1$. I want to show that this sequence is unbounded.
Proof: Suppose that the sequence is bounded. Let $K = \sup a_n $. Then there exists $n \in \mathbb{N}$ a_n such that $$K-1 < a_n < K$$ therefore $$K < \frac{1}{a_{n-1}} + a_{n-1} +1 = a_{n} + 1$$ $a_n + 1 > a_{n+1}$ because $a_{n+1} - a_{n} = \frac{1}{a_n}$ so it is enough to show $\frac{1}{a_n} < 1$ therefore $K < a_n + 1$. Contradiction since $K$ is supremum.
@JohnJack You get $a_{n+1} < a_n + 1$, not $a_{n+1} > K$. It's fixable, but as is, you're not getting where you want. What you need is that $\frac{1}{a_n} > \frac{1}{K}$ under the assumption. That gives you a lower bound for $a_{n+1}$.
@MikeMiller Easy. Don't get up in the middle of the night, stay up until after.
@JohnJack For $n > 1$. For $n = 1$, it depends on the choice of $a$. But, for $x > 0$, you always have $x+\frac{1}{x} \geqslant 2$, so $a_n \geqslant 2$ for all $n > 1$.
@JohnJack Something like that. It doesn't work with $1$, since $a_{n-1} = a_n + \frac{1}{a_n} \leqslant a_n + \frac{1}{2}$ for $n \geqslant 2$. But if $a_n < K$, then ...
@JohnJack For $n = 1$, you have no restrictions on $\frac{1}{a_n}$, it can be any positive number. But since you're looking at the limiting behaviour, you can ignore $a_1$, or "re-index and assume $a_1 \geqslant 2$".
