@copper.hat Sad. Cigarettes be cringe. Cigars are better and don't capitalize on addiction to nicotine.
@user178758 You wouldn't happen to know if there is a similar method of argument reduction for computing logarithms as there is for computing the exponential, would you?
I don't follow. If you're just puffing the cigar, i.e. letting it linger in your mouth, then blowing out, you're not letting tar build up in your lungs or let nicotine directly into the blood stream.
Now, I haven't yet researched it, but it would seem reasonable to me to presume that the exposure to the tar (which is the carcinogen) would be like exposing one's self to the char on your grilled steak.
i had an officemate my last year of grad school who was always quitting smoking. i'd walk into the office and i could tell by the look on her face that she would tear my head off if something upset her.
It's a shame, really. Had the companies existed for their true purpose, to create good products which people need, we wouldn't have such problems like cigarettes and nicotine addiction.
iowa lost to purdue today. purdue! you wanna talk addiction, i wonder what some of those referees were on. iowa still would have lost. they are dead to me now.
as mitch hedberg said. "i want to be a race car passenger. just a guy who bugs the driver. 'say man, can I turn on the radio?' 'you should slow down.' 'why do we gotta keep going in circles?' 'can i put my feet out the window?' 'man, you really like Tide.'"
The real question, though, is how's the beer on the winning car taste after a year and can I get a swab to make that winning homebrew.
I can imagine, but I mean, if I'm being honest, I'm heavily biased against sport because I'm not into it in the least, and so if there's anything to appreciate, then in my totally objective and unbiased opinion, the appreciation requires a microscope to really get.
something like driving, there's a lot of strategy. it's not just go fast because most of those cars can go the same speed, or if they can't there's little the driver can do.
i joke about rowing being dumb because my wife does it. she has some kind of subscription to a rowing magazine. as i understand it, the way to win a rowing competition is to be faster than the other competitors. that's it.
they often don't even race at the same time.
it's moronic that there are magazines about this.
running is pretty dumb too, although as a hobby i understand it. stupid sport.
Like, the car racing thing is just a bunch of cars whose performance appears to vary within margin of error of each other unless it's the more interesting street races with real racing cars as opposed to mere stock cars.
Like the twisty circuits as opposed to the flat circle.
Ok, so I'm assuming it's a matter of ensuring the center of mass is, well in the center of the boat; the alternative is there's weight balancing among teams and they try to normalize the weight that all teams must deal with to make the sport more fair.
@leslietownes You didn't insult them enough. You have to treat them the way you do mathematical proofs: you stare them down until they comply to assert dominance over your adversary.
Tell them their coffee was brewed poorly or smth idk
here's a permutations question. suppose i have a set of permutations on $n$ objects which i know generate the full $S_n$. how would one go about finding the smallest (by number of factors) number of permutations required to generate a particular element?
(in my case of interest, i've got a set of 9 permutations on 8 objects, and there's another permutation I'm trying to write in terms of these 9 as simply as possible. so far the smallest product i know uses 20 factors, and that seems...excessive.)
@copper.hat But I don't need the partial derivative with respect to $t$. I need to prove that the partial derivatives of $p_r(\theta-t)$ with respect to $\theta$ and $r$ are also continuos with respect to $t$.
@Semiclassical what do you mean by "generate the full $S_n$" and "finding the smallest number of permutations required to generate a particular element"?
suppose you want to generate any permutation on n elements. then it's enough to use transpositions, for instance. more precisely, we could restrict ourselves to adjacent transpositions swapping k and k+1
so, for instance, the permutation (1324) on four elements may equivalently be written as (23)(12)(34)(23)
so the transpositions (12), (23), (34) are enough to express any permutation
here's a permutations question. suppose i have a set of permutations on $n$ objects which i know generate the full $S_n$. how would one go about finding the smallest (by number of factors) number of permutations required to generate a particular element?
if you have goofy generators, how does one find minimal representations?
hi, im struggling to apply DCT to show that this map is continuous: Let $R$ be a fixed closed rectangle in $\mathbb{C}$, and let $f \in L^p(R)$ (complex valued) for some $p > 2$, the map is $R \rightarrow \mathbb{C}$ , $z \rightarrow \int_{R} \frac{f(w)}{w-z} dw$
anyone see a way forward?
the issue im having is that i dont know how to majorize $\frac{1}{w - z_n}$ in $R$, where $z_n \rightarrow z_0$ in $R$
basically since $\lim_{n \rightarrow \infty} \frac{|w-z_0|}{|w-z_n|} = 1$ does not happen uniformly in $w$, and we need larger and larger $n$ for $w$ closer and closer to $z_0$ in $R$
sorry, that integral should read $d\lambda(w)$ at the end where $\lambda$ is the two-dimensional lebesgue measure, rather than $dw$ (its not a line integral)
@copper.hat I get it but I don't want to prove that the partial derivatives with respect to $t$ are continuos in $t$, I want to prove that the partial derivatives with respect to $r$ and $\theta$ are continuos in $t$.
i have a hacky method in mind, namely prove this for $f \in C_{0}^{\infty}(R)$ (by rewriting it as $\int_{H} \frac{f(w' + z)}{w'} d\lambda(w')$ where $H$ is some large rectangle containing all of the $R - z_n$ and $R - z_0$ in its interior) using uniform continuity of $f$ (so DCT is easily applicable), and then apply a density argument to get the general result.. but im hoping there is a direct proof
the $p > 2$ here is pretty essential (for the hacky argument) or else we wont get to use holders
In fact, from the manner in which it was solved, it sounds like the definition of NP-complete
I don't know if this is relevant, but if factors here refers to products of composite integers, I've learned a bit in this regard through modulus, but I don't think it is of much use as-is.
I would consider it maybe a sort of gateway or food for thought, but I learned for the most part that you can list factors of a number in the form of $2^n x$ as all power of two factors of $x$ relative to some modulus. The only thing missing then is all the factors in-between.
It's something about equivalence classes and the fact that you can end up getting powers of two from sums and differences in those equivalence classes that you get something easily (relatively speaking) manipulable.
I just thought it might be interesting given that I heard something about factorization of integers not having an efficient algorithm. I believe it was prime factorization or something similar in the same category.
if $p > 2$, and I have a continuous linear functional $L^p(K) \rightarrow \mathbb{C}$ where $K \subset \mathbb{C}$ is a closed rectangle, why is this functional necessarily of the form $f \rightarrow \int_{K} \psi f d \lambda$ where $\lambda$ is the two-dimensional lebesgue measure, for some $\psi \in L^p(K)$?
for $p = 2$ I can see this via riesz representation, but im not sure why it follows as well if $p > 2$
@BannedUser In case it happened to be so, it usually indicates left logical shift traditionally in most programming languages. This is all I have to contribute.
In otherwords, multiplication by $2^x$ for $x\in\mathbb{N}_0$ usually.
@user178758. Thank you. Appreciated. Yep I found that indeed means "mean, median, etc." ~_~
Question please about solving this recurrence relation.
You can compute the indices for the probe sequence more efficiently by using the recurrence relation $k_{i + 1} = (k_i + 2i + 1)$ modulo n for i ≥ 0 and $k_0 = k$ Derive this recurrence relation.
(Law of the iterated logarithm) let $\lambda > 1$. He claims here that I can pick $\lambda > 1$ so big, so that $$ \dfrac{(1-\dfrac{\epsilon}{10})\sqrt{2(\lambda^{k+1}-\lambda^k)\log\log(\lambda^{k+1}-\lambda^k)} - (1 + \epsilon) \sqrt{2 \lambda^k \log\log(\lambda^k)}}{\sqrt{2 \lambda^{k+1} \log\log(\lambda^{k+1})}} \ge 1 $$
which is categorically false, because already $$ \sqrt{2(\lambda^{k+1}-\lambda^k)\log\log(\lambda^{k+1}-\lambda^k)} < \sqrt{2 \lambda^{k+1} \log\log(\lambda^{k+1})} $$
I think I know what he meant to say. and it's not it ^
Assuming you mean modulus, then for $x\bmod y$, $\lbrace x,y \rbrace \in \mathbb{N}$, and $y\leq x\lt 2y$, you can conditionally subtract $y$ from $x$ to compute the modulus. This obviously requires reducing $x$ such that it satisfies the inequality if $x$ as-is does not satisfy it.
The base is $y$ and we want to compute $\frac{1}{y}$.
To convert to base two for an arbitrary amount of precision, you begin with 1 and repeatedly double it in base $y$. Double it once. If the result is greater than or equal to 1, then the corresponding bit is a 1; 0 otherwise. Take the fractional portion of the result and repeat up to $n$ times for $n$ bits of precision.
In case you didn't notice, we only have to double and take the fractional portion which means that for $x$ doublings, the fractional portion is $2^x\bmod y$.
So, find a way to compute $2^x\bmod y$ fast enough and you can compute reciprocals quite quickly this way for each bit in parallel.
Exploiting the properties of commutative rings, then we can represent $2^x$ as the product of several smaller powers of two, therefore, on average, the cost to compute modulus should be O(log n). We wish to do this for n bits, therefore the cost to compute the reciprocal should be O(n log n).
And, obviously, you can compute multiplies if you can compute reciprocals by dividing by the reciprocal of the divisor, so you could probably make something in hardware that computes multiplication and division in the same execution unit.
No clue, honestly. I presume it does assuming you're trying to find an integer in some modulus like you would be with the method presented here: stackoverflow.com/a/48482313/2950711
If you look at the number there which multiplies $x$, it is actually a multiple of $\frac{1}{3}$.
the line of thought i'm considering is, if a is invertible mod n, then gcd-with-bezout will produce P and Q with Pa + Qn = 1 and then P is an inverse of a mod n
I would love to just wait around and let someone else prove this, but I think in the mean time, I'm just going to implement this and hope someone else can prove it for me in the mean time. It is quite reasonable to believe that I should begin implementing this now.
@leslietownes. I am just trying to show like if the module is given by (x+1) mod n, then how we can replace the mod with one subtraction and comparision? 1) check if (x+1)>n is the first step I see, then we can subtract (x+1) - n if the comparision is true
avra if i'm understanding things correctly, in general, you may need iterated subtraction, not just one subtraction. that's the end of it. what is the end game here? is this a textbook problem or an applied problem or what?
as a matter of prose, i don't like "occasional" in that, if an algorithm is designed to subtract, over and over without interruption, until [condition] is satisfied, i would not personally describe the subtraction aspect as "occasional"
i don't have much insight into this problem, it seems poorly written and i am a terrible mind reader
@AMDG I don't see an algorithm there. I see calls to mod, and if you add one of those for each bit of a general number, I don't see that it will be any more efficient than simply repeated shifted subtraction.
perhaps there is more buried in desmos code that I don't see
@robjohn There is more buried in it that isn't immediately obvious. The first iteration of the algorithm which is used to reduce the input, $2^x$ to be $2^{x\prime}\leq 2^x$, is in there as $2^{n}\left(2^{\operatorname{floor}\left(\log_{2}\left(y_{0}\right)\right)+1}-y_{0}\right)-2^{n-1}y_{0}$.
The starred message three up from the bottom is an example of just one iteration of the algorithm.
$2^7$ was reduced to $-2^3$
We can then repeat this process once more, taking into account the number of factors of $-1$ that we accumulate per iteration.
And just plug in, for this example, $2^3$ again to reduce it further, but it cannot be reduced further for $2^x\bmod 11$ since $2^3 \lt 11$, so we now multiply by -1 to get $-2^3$ again and finally compute the modulus as a single subtraction (or addition).
It could probably be reduced to a constant-time implementation if we could efficiently compute addition-subtraction-chains and/or the number of iterations required to reduce $2^x$ which is always a finite number of iterations.
Also I'm quite certain the cost decreases as $y$ increases.
However, is this an iteration that must be performed once per bit of the number being modded? I assume so since you are only doing this on powers of 2.
to apply the mod to a general number, you will need to break it into powers of 2
This doesn't seem more efficient than simply subtracting and discarding the result if a carry is performed.
It's not for computing $x\bmod y$, only $2^x\bmod y$, and that, for determining the value of a single bit for the reciprocal of $y$.
$x$ is the index of the bit we wish to compute for the reciprocal of $y$. Then the value of the bit using Iverson bracket notation is $[2\left(2^x\bmod y\right)\geq 1]$.
From there, just use the resulting division algorithm to compute the modulus of any two values rather than just powers of two.
Oh and of course $x$ is an integer beginning with $0$ as the index of the least significant bit.
Just make sure to compute using fixed point.
So actually it's $[2\left(2^{|x|}\bmod y\right)\geq y]$ for $x\lt 0$.
@AMDG It might only compute the highest order bit for $1/y$ since any lower bits would be masked by other bits of the "numerator". I guess I am just not getting how this computation helps in general.
this is one of the things my analysis instructor blew. there's an exercise in rudin relating to this, the instructor offered a 'simplification' that actually wasn't a simplification.
in German we have two verbs "umfahren" and "umfahren" which have the stress on different syllables. The meanings are basically opposites. If you stress the first syllable, it means "drive over", if you stress the second syllable, it means "drive around"
OOM-fahren. oom-FAHRen. got it. i may need this, i have a german friend who visits sometime, and long beach has a roundabout between our house and the airport.
people would strategically flee from 104/113 if taught by actual faculty. the word on the street was the faculty were 'too hard' and the postdocs were 'easy because they can't fail too many people.'
not true as a factual matter, but true in practice.
Often postdocs are better teachers than tenured faculty. Even teaching as a grad student I was a better teacher than many of the faculty. Yes, and modest, too.
They let me teach my own Guillemin & Pollack course, which then ultimately became a course in the curriculum.
i don't think it's an accident that at many schools, non-tenure-track faculty are disqualified from winning university-wide teaching awards. it would embarrass them.
