I got $\inf\{|s_n|\} + \epsilon > |s_\delta| \geq \inf \{|s_n|\}$ and $|s_\delta| > |s_{\delta + 1}|$. Is there an algebraic way to get $\epsilon > |s_\delta - \inf\{s_n\}|$?
i can do the argument with words but feel less cool
@shintuku MS-DOS was mid-80s. First public version was 1983. And orange and beige?? I don't remember much of that, unless you have a bunch of faded posters.
either increasing, decreasing, nondecreasing or nonincreasing
i think it allows either positive or negative sequences, let me check
from Bruckner et. al.'s Elementary Real Analysis p67. but then again i just picked some random recommendation from MathSE to learn analysis so I don't know if this is this is standard
@shintuku it looks to be on page 47, from the copy I just downloaded.
Thus every increasing sequence is also nondecreasing but not conversely. A sequence that has any one of these four properties (increasing, decreasing, nondecreasing, or nonincreasing) is said to be monotonic. Monotonic sequences are often easier to deal with than sequences that can go both up and down.
oh I wasn't doing any problem in particular, I was doing a particular case of the monotone convergence theorem, one for bounded sequences which are either strictly positive or strictly negative and tend towards the zero line without crossing it, maybe being bounded before touching it
the more general proof given by the author works for this particular case
@Koro hey Koro, thanks a lot for the answer! I saw $|a_n| - l < \epsilon$ as sufficient for a limit, but do you know if it has a relation to the $|a_n - L| < \epsilon$ definition, or is it actually not derived but also an acceptable alternative definition?
@shintuku: you're welcome. To the comment I made in the answer, I will also like to state that $||a_n|-|l||\le |a_n-l|$ is always true. I hope your doubt is clear now :)
@shin, note that we also proved one more thing in the answer: that is, $L=\inf (|a_n|)$ if $(|a_n|)$ is a strictly decreasing sequence.
and therefore $L\ge 0$ and for any $n$, $|a_n|-L$ is non negative. :)
and this allows us to say that $||a_n|-L|\lt \epsilon$ and hence $|a_n|$ converges to $L$. :)
if $a_n$ is monotonic then eventually it will stop changing signs (+,-) and have the same sign for all large N terms.
And hence for large n, $|a_n| $ will either be $a_n$ for all large n or $-a_n$ for all large n. And since convergence of a sequence won't change by multiplying by minus sign, the result is proven!
if you have $\mu E = \int 1_E h d |\mu|$ for all $E$ you have $\int f d \mu = \int f h d |\mu|$ and so, letting $f = {g \over h}$ you have $\int g {1 \over h} d \mu = \int g d |\mu|$ for all $g$ and so letting $g=1_E$ you have the desired result.
We are given 9 numbers : -6 , -8 , -5 , 10 , 6 , 4 , 8 , 1 , 9 , 7 Q is to write an algorithm for this such that the addition of any two numbers always give 2 .
you could also notice that the real & imaginary parts of $\mu$ are absolutely continuous with respect to the total variation hence Radon Nikodym applies. then show that the combination of the derivatives is ${ 1 \over h}$.
@shintuku not always. Take $a_n $ as the sequence $1,-1,1,-1,...$. Particularly, for you question: I request you to refer my last detailed comment above. In case of monotonicity we have certain advantage :)
shintuku, if you repeat the arguments for the case when $(|a_n|)$ is an increasing sequence, then you'll understand it better :)
Let $f(x)=\begin{cases}1 \text{ if }x\in \mathbb R\setminus \mathbb Q\cup \{0\}\\ 1-\frac 1p \text{ if } x=\frac np, n\in \mathbb Z\setminus\{0\}, p\in \mathbb N \text{ and gcd}(m,n)=1 \end{cases}$
The points at which $f$ is continuous, are to be found.
Let $a\in \mathbb R$ be any point in $\mathbb R$. So I considered the following two cases:
$1)$: $a=\frac np\in \mathbb Q\setminus\{0\}$
Let $(x_n)$ be a sequence of irrationals converging to $a$. For continuity at $a$ we must have $\lim f(x_n)=f(a)=\lim 1=1=1-\frac np, $ which is not possible. Hence, $f$ is not continuous at any …
@shintuku you can't simply write $\inf|a_n|$ without telling what the set you are taking the infimum over. $\inf\limits_{n\in\mathbb{N}}|a_n|$ is at least a quantity.
Let $\{a_n \}$ be a monotonic sequence with $|a_n| > |a_{n-1}|$ with both an infimum and a supremum. Since $\inf\{|a_n|\} + \epsilon$ is not an infimum of $|a_n|$, there exists an $N$ s.t. $n > N$ s.t.
"Since $\inf\{|a_n|\} + \epsilon$ is not an infimum of $|a_n|$..." - It would be more correct to say "Since $\inf\{|a_n|\} + \epsilon$ is not a lower bound of $|a_n|$..." After all, $\inf\{|a_n|\} - \epsilon$ is also not an (the) infimum, but the rest wouldn't follow in that case.
@shintuku If $\{a_n\}$ is increasing, then for all $n$ we have $a_{n} \leq a_{n+1}$. By assumption we also have $|a_n| > |a_{n+1}|$ for all $n$. It follows that $|a_n| > |a_{n+1}| \geq a_{n+1} \geq a_n$, hence $a_n < |a_n|$, hence $a_n$ is negative for all $n$. Therefore $a_n$ converges to $\sup\{a_n\} = \sup\{-|a_n|\} = -\inf\{|a_n|\}$.
Similarly, if $\{a_n\}$ is decreasing, then for all $n$ we have $a_n \geq a_{n+1} \geq -|a_{n+1}| > -|a_n|$, hence $a_n$ is positive for all $n$. Therefore $a_n$ converges to $\inf\{a_n\} = \inf\{|a_n|\}$.
Let $f(x)=\begin{cases}1 \text{ if }x\in \mathbb R\setminus \mathbb Q\cup \{0\}\\
1-\frac 1p \text{ if } x=\frac np, n\in \mathbb Z\setminus\{0\}, p\in \mathbb N \text{ and gcd}(m,n)=1 \end{cases}$
The points at which $f$ is continuous, are to be found.
Let $a\in \mathbb R$ be any point in...
@Koro Based on a quick skim, your answer looks correct. At nonzero rational points $x$ the function value is strictly less than 1, so any sequence of irrationals converging to $x$ shows that the function can't be continuous there.
At irrational (or zero) values of $x$, the function value is 1, and in any neighborhood of $x$ there are only finitely many rationals with denominator less than any given N, so any sequence of rationals approaching $x$ must eventually have arbitrarily large denominators.
Well you have $f(0) = 1$, and obviously $f(x) \leq 1$ everywhere. Suppose you want to find a neighborhood $N$ of $0$ such that $f(x) > 1 - 1/n$ for all $x$ in $N$. What if you take $N$ to be $(-1/n, 1/n)$? For all irrational $x$ in this neighborhood, of course we have $f(x) = 1$. If $p/q$ is a rational in this neighborhood, after canceling common factors from $p$ and $q$, you must have $q > n$. Therefore $f(p/q) > 1 - 1/n$ as desired.
@Koro (Sorry, too late to edit, I should have said "if $p/q$ is a nonzero rational in this neighborhood...")
@Bungo Thanks a lot. I was thinking along the similar lines while writing my post. I take my post as correct now. Thanks a lot. I would like to mention that first I considered three cases (third case being a=0) but then I noted there was no need for that.
@copper.hat I don't think this works. Could you provide more details? Lebesgue's MCT, and approximation of measurable functions by simple measurable functions - I've seen all these things for positive measures, and positive functions. Here, the measure is complex and the functions are too. Thanks!
i don't know why i can't sleep. i'm more awake than i am in the daytime. i used to live like this but when i had a kid i adjusted to a normal schedule.
and now i'm up again.
my cat's looking at me like, "why are you awake?"
this is her time to prowl the house and do what she wants to do, and here i am being awake and ruining her freedom.
i held her for about an hour trying to go back to sleep and her purring did relax me but i'm more awake than not. today will be very interesting. i have one appointment at 11am. after that i can probably go to bed.
olivia was purring so loudly that she woke my wife up.
my wife said, "what the f--- livvy why are you being so loud." then back to sleep.
i love cats because whenever they sleep they arrange themselves in a cute position. they are artisans of comfort.
@PM2Ring Also, if I understand your code directly, your 9-iteration newton method approximation will take 72c plus a conditional jump, though that seems to just be a precision cut-off, so I'll pretend that isn't there. If that's the minimum precision for getting 64-bit reciprocals or quotients, you're 31c slower than div and 56c slower than fdiv for integer divide (15c fdiv, 1c cast).
@leslietownes It's better than C++ and I'd rather deal with memory management than the dreadful, absolutely horrific and positive evil of garbage collection that destroys performance.
one time my wife was like, what are you and your best friend talking about all day. i gave her the phone. about 100 texts about celebrities. we are stupid and hollow people.
"But I have pointers everywhere and that's error prone!" Cool, make a heap or memory manager to manage your pointers and make sure it checks for NULL so that even if you free when you shouldn't, you don't deallocate memory that isn't yours, but instead warns you that you've freed when you shouldn't. Simple solution to a simple problem.
@OliverDiaz Could you check if this (mathb.in/61171) is right? It's my complete attempt for the problem we discussed yesterday. The only gap is whether $$\color{red}{ \int e^{-int}\, d|\mu(t)| = \int \frac{e^{-int}}{h}\, d\mu(t) \quad\quad \text{Is this true?}}$$ is true, i.e. if $d\mu = h d|\mu|$, then can we say $d|\mu| = \frac{1}{h} \cdot d\mu$? I have colored this in red in the solution. Please share your thoughts - thank you!
@copper.hat Tagging you so you can take a look at mathb.in/61171 too. It's the same problem, Rudin's 6.7.
@leslietownes Agreed, but what's even more powerful is a "high-level assembler" that gets it right the first time instead of doing a hacky patch-job that sometimes introduces language bugs and is unapologetic about breaking existing standards if necessary to get a better system that can be easily adopted and migrated to.
i was not being offered enough money to make a future sensible. the landlord wanted one thing, the university was offering another. it was purely financial.
i responded to my last offer with, i literally cannot do this. i wasn't negotiating, i know they didn't have room to negotiate.
a lot of university jobs in the USA are in places with higher than average costs of living.
@leslietownes Makes sense. Do you think it's like that for almost everyone who stays in math academia? I'd like to do a PhD and get an academic job in the US, but the financial aspect always bothers me LOL
there's a school of thought where finances don't matter. i'm of the school of thought where finances come first. as you note it is very hard to transition back to academia from somewhere else.
True. You'd be surprised though, I was just looking up some statistics, and I saw that PhD students in Switzerland are paid 35-40k USD more than those in the US, per year
but it's not impossible. i think it's helpful to maintain relationships with academics while you go elsewhere so it doesn't seem like there is a disruption in your output even if you are technically not in academic.
i talk with my german friend about this a lot. she was actually earning money while a phd student. i zeroed out at the end of the year. whatever came in had to go out immediately.
her state also pays for her child care, which i have to pay for myself.
i can't decide which one of us is radicalizing the other.
she isn't married yet. i keep telling her boyfriend to man up and seal the deal but what do i know about europeans.
maybe there is a tax advantage to remaining single, who knows.
yeah, they are apparently just throwing money around all over the place.
i am very nervous about the political climate in the united states, europe might be a better deal.
my wife was trying to find jobs in europe for a while but gave up when she got tenure.
i have a friend in korea who wants me to join his startup but the idea of moving my entire family to a place where i don't speak the language and know maybe three people seems like a bad idea.
I want to find no. of bijections $a:\mathbb N\to \mathbb N$ such that the series $\sum_{n=1}^\infty\frac{a(n)}{n^2}$ converges.
For brevity, let $a(n)=:a_n$ and now the necessary condition for convergence of the series is $\frac{a_n}n=o(n)$ and hence for large $n$ we should have $\frac{a_n}n\lt \...
it was supposed to be. i asked a guy who worked on von neumann algebras for his whole life and he said "Yeah i don't know anything about that." that was approximately when i stopped trying to prove it.
@robjohn sorry about that. I got confused. It will however be correct to say that "conditionally cgt. series can be rearranged to give us any desired sum" or more generally
a rearrangement that gives us a desired limsup and a desired liminf of partial sum of the rearranged series.
@robjohn I am going through this inequality, I didn't know it before.
that's a very powerful inequality. Thanks a lot. I never knew it before.
But I still don't understand how they got $\sum \frac n{n^2}$ in the answer. We say let $y_r=\frac 1{(n-r+1)^2}$ and clearly $y_1,y_2,...$ is increasing. There is no reason to believe that $\{1,2,3,...,N\}\subset f[\{1,2,3,...,N\}]$ for all large N
No, the $\frac1{n^2}$ is decreasing and $n$ is increasing. The rearrangement inequality says that $\sum_{n=1}^\infty n\left(\frac1{n^2}\right)$ is minimal.
Given a bijection that is not increasing, you can find one that has a smaller sum by swapping any two values which are not ordered.
You don't need a finite sum for the Rearrangement Inequality
@AMDG Which parts of that code can't you read? Most of it is very close to C (apart from the lack of semicolons & braces). Eg, the frexp & ldexp functions behave just like the C functions of the same names from <math.h>. The loop for i in range(9): is equivalent to for (i=0; i<9; i++).
I used a Python parallel assignment to handle the sign of the input arg. yy, s = (y, 1) if y > 0 else (-y, -1) is equivalent to:
if (y > 0) {
yy = y;
s = 1;
}
else {
yy = -y;
s = -1;
}
Instead, I have to assume what the data type is from the operations of which they are not making it very obvious since you can add floats and you can add ints as well. Can't tell if there's any implict casting, etc. I'm sure you get my point on why this isn't too readable for me. :)
@AMDG In that code, most stuff is double precision floating point, apart from s, e, and the loop counter i.
Comments in Python start with #. So that middle section is mostly commented-out. They're just alternative polynomials that estimate the reciprocal in the [0.5, 1) range.
Hello I need some advice that I need for analyzing my data but i'm not sure how to approach it
*If group A has 1,000 people and group B has 200, then we must find a measure wherein both groups can be compared to a similar scale. This can be a measure such as an average. The problem with using an average is that outliers can play a huge role in this metric and can misrepresent the general behavior of the population. But removing this outlier can be misleading because we also want to measure the overall effect of the entire group (not just the general behavior).
@PM2Ring Ah. In which case that's slow. Your code, not even considering the fact that the Python implementations that I've seen (or heard of for that matter) are interpreted, would be over a hundred cycles. I can probably compute FSINCOS faster. If there's a fast integer arithmetic with double data type hack available, then that's fine.
That is of course assuming the compiler doesn't automatically vectorize the loop of which there is no guarantee which is why C and assembler are what I use for now.
@AMDG That loop has a limit of 9 but it never goes that high. It usually breaks out of the loop in 3 or 4 steps because Newton's method converges quadratically. i.e., the number of correct bits doubles on every loop. And the initial approximation is already pretty good.
How many iterations are needed for 64 bits of fixed-point mantissa?
I mostly only care about O(1) implementation with same cost for every case, so however much is needed for qwords will also be the cost for bytes, words, and dwords.
@AMDG Why is Python's running speed relevant here? I'm not suggesting that you should be writing your code in Python! I just wanted to show you an algorithm, and to give you a way to run it so you can see how well it works.
Sorry, hard not to rant about these things sometimes :)
But again, that 72c is assuming that I compiled your python code to x86/x87 in my head, but you said it's doubles, so that will be significantly more than 72c.
fadd alone is 5c, 1 throughput.
I can get 1c, 4 throughput with fixed-point add since it uses integer arithmetic.
(Which, honestly, makes me beg the question of why x87 arithmetic is so expensive. I don't think I'll ever use float instructions except in the case where it happens to be faster [somehow] than using integer arithmetic)
So I'm sure you can see why Newton's method isn't a terrible idea, but also likely not better than using hardware divide.
The single bottleneck I have for finding a subpattern within a pattern is a single divide, and the other instructions take 6 or 7 cycles, but we'll still be waiting 34c for the divide to catch up. Not to mention, having a fast divide implementation also opens up algorithmic possibilities that would previously have been considered impractical.
The best lead I have right now is solving the knapsack problem for computing reciprocals, so that's what I'm working on.
@AMDG You can use that algorithm with your own fixed-point numbers, if you want. FWIW, in my early days of coding, the CPUs only did integer arithmetic. All floating-point stuff was handled by library code that did everything using integers.
That initial reciprocal approximation has minimum error for a linear polynomial in that range. Its maximum error is sqrt(2) - 3/2 ~= 0.08576, which occurs at the endpoints of the range and at sqrt(1/2). So it doesn't take many loops to get 53 bits of precision (the maximum for IEEE-754 double-precision floating point).
If $\mathrm{G}$ is the centroid and $\mathrm{I}$ the incentre of the triangle with vertices $\mathrm{A} \equiv(-36,7), \mathrm{B} \equiv(20,7)$ and $\mathrm{C} \equiv(0,-8)$, then find $\mathrm{GI}$
Is there any alternate way to the regular method of using centroid and incentre formula?
Still, for 53 bits of mantissa, 4 iterations would be about 32c which is better than 41c, so I'll add that to the list of solutions, but I doubt that unless it can be parallelized, I will not be able to use it.
I currently have an idea right now as well for the knapsack problem. I realized that if I have 1/5 = n, 1 = 4n/5 + n/5. I can choose a power of two numerator and maybe work something out to end up with 1/5.
1/5 is just an example of course that I'm using to work this out.
The quadratic initial approximations give you more starting bits, and that can reduce the number of iterations, which may be useful if you want more than 53 bit precision. I'll post some code that lets you compare the graphs of the various approximation functions. You can see the functions themselves, or select the "flat" option to see f(x) - 1/x
FWIW, my answer here: math.stackexchange.com/a/1295561/207316 has C code (by Dik Winter) that calculates hundreds of decimal digits of pi, using only integer arithmetic. It also has Python code I wrote to compute large numbers of digits of e, mostly using integer arithmetic. It uses some floats to compute optimal loop sizes, but that's not strictly necessary.
I don't fully understand Dik Winter's C code. I assume it's computing some Taylor series. I did a Python version of it a little while ago that can compute as many digits as your RAM allows, because it uses Python integers, which can grow to arbitrary precision.
@AMDG It was designed to be difficult to read. ;) It's an example of "code golf". There's a whole Stack Exchange site dedicated to that sort of thing. And before that, there was the IOCCC en.wikipedia.org/wiki/International_Obfuscated_C_Code_Contest
I've posted a few answers on codegolf.stackexchange.com but the prominence of special-purpose golfing languages takes a lot of the fun out of it for me.
@Pherdindy First, make the code work correctly. Then optimize any bottlenecks, if necessary. Of course, it's a Good Idea to have some familiarity with the complexity of the algorithms that you're using so that your initial solution doesn't do really slow stuff. ;) Knuth famously said that 97% of the time, premature optimization is the root of all evil. Yet we should not pass up our opportunities in that critical 3%.
@epsilon-emperor @epsilon-emperor: The first argument is not quite right. Remember that the quantities involve are complex; hence the relations of the form $a+ib<c+id<e+if$ are not valid (the order from $\mathbb{R}$ does not carry over nicely to $\mathbb{C}$).
@epsilon-emperor: What you do have is that of $p$ is a trigonometric polynomial that approximates $f\in C(T)$ bat say $\varepsilon>0$ in the uniform norm, then $\Big|\int_T (f(t)-p(t)) e^{-int}\,\mu(dt)\Big|\leq\int_T|f(t)-g(t)||\mu|(dt)\leq\varepsilon|\mu|(\mathbb{T})$
Can someone explain whats going on, Im reading something which says the lipschitz constant of the function $\rho_t+V : \mathbb{R}^n\to \mathbb{R}$ is decreasing with time, and that : if $V$ is uniformly m-convex i.e $DV\geq m I $ then $e^{mt} Lip(\rho_t + V)$ is also decreasing
is this a valid method to find the middlepoint of a circle? Describing a tangential line is not possible for the project i make atm. can only intersect. ibb.co/FgRk9Xj
is y always the diameter if it's at 90° from a secant line?
Suppose I have 2 linearly independent vectors $a$ and $b$. $a$ and $b$ are both $n$ dimensional. and $a$ and $b$ are linearly independent. This means that $a^Tb$ is not equal to either norm a ^2 or norm b ^2(By Cauchy Schwartz) Does this mean $\|a\|_2/(\|a\|_2 + a^Tb) is independent of n. That is $a$ is O(1)
@OliverDiaz I'm reading it. So about the first blue section, I understand your argument for all $f\in C(T)$, but I do not see why my argument fails? The one with $\liminf$ and $\limsup$?
I basically showed that $0\le \liminf \le \limsup \le 0$, so $\lim = 0$
@epsilon-emperor: Your argument is not entirely kosher since the quantities involved are complex, inequalities are not valid here: $a_n+ib_n < c_n+ic_n$
