Mathematics - 2023-08-15 (page 1 of 2)

Xander Henderson

00:00

@Jakobian Curse you, and your sons, unto the seventh generation.

Jakobian

there's a cover of Teen spirit by Patti Smith that I like
but it's not the greatest song by Nirvana or anything

leslie townes

i mean, it came out over 30 years ago. you wanna feel old, xander? "hey ya!" came out 20 years ago and is also an oldie.

Jakobian

@XanderHenderson what, I was just wondering if old people listen to cool bands too

:P

Xander Henderson

@Jakobian ...unto the TENTH generation.

冥王 Hades

Has it been 243 years yet? Time for me to reincarnate?

Jakobian

00:03

it's okay because I won't have children
I hope the curse doesn't accumulate then

leslie townes

when i was a kid, "oldies" meant, like, doo-wop stuff that people slightly older than my dad would have listened to as teenagers, but not 60s rock that my dad would have listened to as a teenager.

and that's the only right definition of 'oldies.'

冥王 Hades

those hit by my sword are cursed for an eternity to forever be trapped between the world of the living and the world of dead

Xander Henderson

@leslietownes Hear hear!

Jakobian

https://music.youtube.com/watch?v=Ba_08WWIWV8
All Apologies by Nirvana

I like this better than Teen Spirit

@XanderHenderson do you like Queens of the Stone Age?

here's a great song by them music.youtube.com/watch?v=ma5hQm-IVD8

Xander Henderson

00:27

@Jakobian I am not familiar with them off the top of my head. But I appreciate the song to which you have linked. It is definitely within the general sort of music I like to listen to when I have the time to listen.

Jakobian

oh.. well they're a little less known but they've been on the radio I think

if you ever heard of kyuss, they are basically successors of that band

kyuss is considered the origin of stoner rock

Stoner rock

Stoner rock, also known as stoner metal or stoner doom, is a rock music fusion genre that combines elements of doom metal with psychedelic rock and acid rock. The genre emerged during the early 1990s and was pioneered foremost by Kyuss and Sleep. == Characteristics == Stoner rock is typically slow-to-mid tempo and features a heavily distorted, groove-laden bass-heavy sound, melodic vocals, and "retro" production. Due to the similarities between stoner and sludge metal, there is often a crossover between the two genres. This hybrid has traits of both styles, but generally lacks stoner metal's laid...

I think... over the internet at least, stoner rock is pretty popular genre

GratefulDisciple

00:48

@冥王Hades I used to work for a hard drive manufacturer and knows about how they do support and data recovery. Opening the "casket" is a big no no. You really shouldn't do that yourself. You have most likely ruined the remaining chance of recovery.

@冥王Hades SSD has its own issues and vulnerabilities. I haven't read about whether there is recovery from the NAND chips or whether they can switch out only the controller portion if that's what's malfunctioning.

As @XanderHenderson says, the moral of the story is: proper backup. Commercial companies do the 2-2-2- rule: two sites, two copies, two media types. That's gold standard but it's overkill for us.

冥王 Hades

@GratefulDisciple the only immediate problem with NANDA chips AFAIK are their limited read and write. But by the time I even approach that limit I’m on a new SSD

@GratefulDisciple yeah I knew I shouldn’t really take it apart myself but….curiosity gets the better of me every time. To be fair it’s hardly the first (or last) time that my curiosity has cost me dearly

GratefulDisciple

@冥王Hades I have been using Samsung SSD for years. Very happy with them. They have a utility showing the TBW (total bytes written) as a measure of the usage, and have a smart firmware that can automatically move problem sector to another one by designating a small percentage as spare. Windows also help in reducing unnecessary writes.

(sorry TBW = Terabytes Written).

Xander Henderson

01:13

@GratefulDisciple Yeah, I know the commercial standard. In addition to my computer's internal drives, I have external drives for backing up each device, and I keep all of my important files on GoogleDrive and DropBox. Not quite 2-2-2, but as close that that as I am going to get on my income.

MathCrackExchange

0

Q: Maximal estimate of how many primes are contained in $(p_i\Bbb{Z} + 2) \cap I_n$ where $I_n = [p_n + 1, ..., p_{n+1}^2 -3]$?

The number of primes in the interval $I_n:=[p_{n} + 1, p_{n+1}^2 - 3]$ is given by: $\pi(p_{n+1}^2 -3) - \pi{p_n + 1 - 1} = \pi(p_{n+1}^2 - 3) - n$. Since by the PNT we have that $p_{n+1}^2 - 3 \sim_{\textbf{is asymptotic with}} (n \ln n)^2 -3$. So that since $\pi(n) \sim \dfrac{n}{\ln n}$ we ...

Yes, prime number madness

GratefulDisciple

01:31

@XanderHenderson Yes, we're not as rich as @冥王Hades 😀. I'm also doing backup similarly.

Jakobian

02:19

in the variety of vector spaces, polynomials can be though of as maps of the form $(x_1, ..., x_n)\mapsto a_1x_1+...+a_nx_n$

so this analogy between polynomials of vector spaces and polynomials of fields gives an even wider analogy between linear independence and algebraic independence

Xander Henderson

02:35

@Jakobian Years and years and years ago, I took a class on numerical analysis, with a lot of coding (in C and FORTRAN and no other languages). A polynomial was an object which contained an unsigned int (degree) and an array of floats (coefficients). Polynomial evaluation was done with Horner.

So $(n, a_n, \dots, a_0)$ mapped to $a_0 + x(a_1 + x(a_2 + \dotsb x(a_n)\dotsb ))$.

Or however that works out. Too tired to math güd right now.

Thomas Finley

@robjohn I am writing this in reference to the answer you gave in my last post. First of all, thank you! But, I think you have made an erring assumption in your assumption. If you have time, please do check the comment I made there.

The place where you assert that $46300.00$ has $7$ significant digits is incorrect, as the $0's$ in the case, are used for "positioning" the decimal point and the zeros that are needed to fix the decimal points are insignificant. Say for example, in $0.0045, $ the digits $4$ and $5$ are the only significant digits. The $0's$ here, are simply used to fix the decimal point so, they are insignificant. You can find this rule in the OP, where I quoted a portion of the book and this case is readily handled in there. — Thomas Finley 6 mins ago

(The above is the comment in totality)

Xander Henderson

I don't see an error in robjohn's reasoning.

Thomas Finley

@XanderHenderson Did you read my comment? According to the definition that's given in my book, his reasoning seems to be a bit inconsistent

Xander Henderson

Though I also think that the whole topic of sig figs is (a) not really about mathematics (it is about physics and chemistry and biology and the sciences in general), and (b) completely cleared up by using scientific notation (where you can be precise about sig figs).

@ThomasFinley Do you know how rude it is to start a reply by suggesting that your interlocutor hasn't read the thing to which they are replying?

Significant figures are a tool for explaining how precise your measuring instrument is. If I have a tool which is accurate to the nearest foot, and I measure the height of a mountain peak to be 14,000 feet, then the number 14,000 has five significant figures (the zeros here are not simply being used to fix a decimal point---they have meaning). If my tool is only accurate to the nearest 1000 feet, then the number 14,000 only has 2 sig figs.

But there is no way to tell the difference between 14,000 (5 sig figs) and 14,000 (2 sig figs) based on the way they have been reported here. Which is why scientific notation is so helpful.

Thomas Finley

@XanderHenderson Thank you for sharing your viewpoint. I didn't mean to be rude in my previous comment. My intention was to discuss the inconsistency I noticed based on the definition in my book. I appreciate your input and would be interested in hearing your thoughts on the matter.

Xander Henderson

02:50

$1.4\cdot 10^{4}$ has two sig figs. $1.4000\cdot 10^4$ has five sig figs.

Thomas Finley

@XanderHenderson sure.

Xander Henderson

I don't see the inconsistency between your book's definition and what robjohn said.

leslie townes

one of my friends taught college algebra (maybe not the exact name of the class, it was maybe 'quantitative reasoning') out of a book that used a convention where when they taught this material, they put the last sig fig in boldface, or underline, or maybe dotted it. basically one of those things you can do to denote a vector. imagine having to teach out of that

"it's like scientific notation, except it's made up just for this class and nobody uses it"

Xander Henderson

Yeesh...

leslie townes

i guess it allows you to cover that before you cover exponents, but yeeesh

Thomas Finley

02:57

@XanderHenderson So, all in all you suggest 24300.00 has 7 sig figures , if I understand you correctly? Then according to your reasoning, 0.007 has all the digits significant. But that is again weird

Xander Henderson

@ThomasFinley Where did I say that?

Thomas Finley

@XanderHenderson Robjohn said, it, and you said, "I don't see the inconsistency between your book's definition and what robjohn said."

Xander Henderson

@ThomasFinley No, that is not how I read robjohn's answer.

Jakobian

@XanderHenderson that sounds right

this is horner's polynomial form or something

I read something about it once or twice

Xander Henderson

As I read his answer, 24300.00 definitely has seven sig figs. But 0.007 may have anywhere from one to four sig figs, and you might assume that it has four if it is in a context with numbers like 1.428.

Thomas Finley

03:00

@XanderHenderson Do you agree that in 24300.00 the zeros are used to fix the position of the decimal point?

Xander Henderson

@Jakobian Horner is good, because the number of multiplications is $n$ (where $n$ is the degree of the polynomial), while the number of multiplications in the "usual" form goes like $n^2$. (in both cases, up to a constant).

@ThomasFinley Yes.

And?...

Thomas Finley

@XanderHenderson According to my book, it's given, "A significant figure is any one of the digits $1, 2,3,... 9;$ and $0$ is a significant figure except when it is used to fix the decimal point or to fill the places of unknown or discarded digits. "

Xander Henderson

I would assume that 24300.00 has seven sig figs, yes. Because if it only had two, or five, or whatever, why bother writing the two zeros to the right of the decimal point. But 0.007 could have four, or one, or anything in between, because you can't tell if the extra zeros are significant, or are just being used to fix the decimal point.

@ThomasFinley Yes, and?

Thomas Finley

@XanderHenderson So, the zeros are not significant in here

Xander Henderson

@ThomasFinley Oh, wait, hang on... in 24300.00, the zeros are not "just" being used to fix the decimal point.

If they were just being used for that purpose, you wouldn't write the extra zeros to the right.

Sorry----I'm tired, and not processing well.

Thomas Finley

03:04

@XanderHenderson we're on the same boat.

Xander Henderson

In the number 24300, you can't immediately tell if the zeros are significant or not, but if one bothers to write a decimal point (and zeros to the right of that point), the general assumption is that those extra zeros are significant.

But, again, this isn't really a mathematics, question, but a question about clear communication in the sciences.

And all of this is resolved by just using scientific notation. Then you don't have to try to read the mind of the author.

Thomas Finley

@XanderHenderson Why not? When we say, £60.00 don't we give £60 ?

@XanderHenderson yes

@XanderHenderson I agree

leslie townes

in real life you switch ink colors at the last significant digit

Xander Henderson

@ThomasFinley I don't understand the problem.

Thomas Finley

@XanderHenderson This is where my problem is. Isn't £60=£60.00 ?

Doesn't matter whether I wrote, 60.0, 60, 60.0000, 60.00.

They are all simply 60

Xander Henderson

03:08

This isn't a math question. This is a question about how real people perform real computations with money. All transactions are assumed to be to the nearest penny, unless someone says otherwise.

Because our "measuring tool" for financial transactions is one penny.

And we all know that.

leslie townes

four wheel drive? how many digits on that?

Xander Henderson

@leslietownes Four and a half.

leslie townes

when you say there are four significant figures, how many significant figures in the "four"

plz

Xander Henderson

Because of the spare, but it is one of those undersized, drive-under-50-mph-until-you-can-replace it spares.

leslie townes

i love those things

Xander Henderson

03:09

@leslietownes One thousand seven hundred ten.

Thomas Finley

@XanderHenderson Say, I wrote a number, 24300.00, in front of you. What will you think about this? How will you perceive it?

Xander Henderson

@ThomasFinley Again, the whole idea of sig figs is to help us clearly communicate the precision of the tools we are using to measure phenomena. This is very useful in the sciences. Not so useful in the more abstract parts of mathematics (where we prefer to work with exact values) nor in "real life", where we tend not to ever require this kind of precision.

And in the sciences, if you don't want to confuse your reader, you use scientific notation to clearly communicate the precision of your measurements.

@ThomasFinley "How will you perceive it?" With my eyes.

Thomas Finley

@XanderHenderson I was having a tough time, understanding this, because this is taught in a section called numerical analysis in our Mathematics Honours course

@XanderHenderson I am helpless here, I'm sorry.

Xander Henderson

@ThomasFinley Yes, because physics departments have pretty universally off-loaded much of their teaching to mathematics departments, so we end up having to teach this nonsense, even though it has nothing to do with mathematics.

Again, the whole idea is that sig figs are supposed to tell the reader how good your tool is. If someone writes 23400.00, then they are claiming that their tool is good to the nearest hundredth of a unit, i.e. there are seven sig figs.

But 24300 is ambiguous, because we don't know if those zeros are there to pad the number to the decimal point, or if they are part of the precision of the tool. I don't know how many sig figs this number has, unless it is in context with other numbers.

Thomas Finley

@XanderHenderson maybe, but I will again say this is controversial as well, unless, you are in the mind of the person who wrote it...

@XanderHenderson I agree

Xander Henderson

03:18

@ThomasFinley No. 24300.00 is not ambiguous in technical writing.

It has seven sig figs.

Thomas Finley

@XanderHenderson Honestly, I don't like physics.

@XanderHenderson "in techincal writing"

Xander Henderson

Again, note that the whole idea of sig figs only makes sense in technical writing. It doesn't apply to casual conversation, or informal writing.

Thomas Finley

@XanderHenderson ok

Xander Henderson

And, again, if it matters, you use scientific notation.

Thomas Finley

@XanderHenderson I am not the one bringing up the question, the book is.

Xander Henderson

03:21

@ThomasFinley No one said that you were wrong.

But your persistence in ignoring everything that robjohn and I have said is a little frustrating.

Thomas Finley

@XanderHenderson I understand what you were trying to say i.e if a number is of the form, say, 700.00 then it has 5 significant figures as the instrument used for measuring the quantity is correct upto two decimal places and so, although the zeros are used to fix the decimal point position but also, a part of the actual measurement so, they have to be significant

Can anyone help me, regarding this assertion?

I feel the assertion is correct, no problem with it. But I think the absolute error is always $10^{-(n+1)}$ in all cases

Xander Henderson

That isn't an assertion. It is a definition. It is description of what the last sig fig actually means.

If I have a ruler which has marks every meter, then I can only report the length of an object to the nearest meter. So if I say that I am 2 meters tall, this means that I am somewhere between 1.5 and 2.5 meters tall. I could be off by as much as half a meter (i.e. $1 \cdot \frac{1}{2} = 0.5$ meters).

Then I get a better ruler which measures to the nearest centimeter. I measure myself again and determine that I am 1.80 meters tall. In this case, I am saying that my height (reported with three sig figs) is accurate to the nearest 0.01 meters. The maximum absolute error is $\frac{1}{2} \cdot 0.01 = 0.005$ meters (or half a centimeter, or five mm).

In any event, it has been a very long day, I am very tired, and going to bed.

Thomas Finley

03:55

@XanderHenderson just one question: Which book did you refer for these understandings?

I need a good book focussing on these things.

PM 2Ring

05:42

Spooky. Someone just posted about Buridan's Ass on Physics.SE physics.stackexchange.com/q/776174/123208

@ThomasFinley We have numerous posts about significant figures on Physics.SE, eg Rules of significant figures, precision, and uncertainty, and links therein.

Thomas Finley

05:59

0

Q: Problem in Understanding the Concept of Accuracy in Numerical Analysis

I was studying about the concepts of Relative , Absolute and percentage errors in Numerical Analysis. I was reading from the book, "Numerical Mathematical Analysis" by J Scarborough. There was a section titled " Relation between Relative Error and the Number of Significant Figures ". I am having ...

numerical-methods significant-figures

Guys, I need a little bit of help with this!

Chiho

10:33

Hello all.
I'm a little bit confused about something.

Two matrices(A,B) are said to be similar if there exists some invertible matrix P such that
A = P-¹BP

But matrix multiplication is associative and that means A = P-¹PB.

But then P-¹P = I (identity matrix)
And therefore A = IB , meaning that A ≈ B (equivalence)

If this is so why does B have an egienvector of PX and A an egienvector of X for some X ≠ 0 and for a shared egienvalue k which both matrices have.

It doesn't make sense to me since both matrices are equivalent or maybe I'm missing something.

robjohn

11:03

@Chiho You should try to install ChatJax and use it here. Matrix multiplication is associative, but not commutative; so $P^{-1}BP\ne P^{-1}PB$.

Chiho

11:32

@robjohn Ah yes I mistook the the two properties and it led to this. I've been cleared up thanks. And yes I'd install ChatJax :)

冥王 Hades

11:59

The Euphrates will dry up before I do well on any history exam (got my grade for it today and I failed)

CowperKettle

@冥王Hades I'm sorry to hear that

冥王 Hades

Eh I’m not that upset. Still aced mathematics exams so I’m happy

CowperKettle

I loved history.

I finished school in 1995, so we didn't even have official textbooks printed for Russia and not for the USSR. The history teacher was nice, she taught us a lot of stuff not covered in the Soviet texbooks.

PM 2Ring

@Jakobian I just listened to this clip of Patti's version of Teen Spirit youtu.be/_QNzQILETLM It's quite good, and I think Kurt would have liked it, and felt honoured to have his song enhanced by Patti's poetry. I don't like everything Patti Smith does, but I've been a fan since the 70s, and still listen to her stuff regularly. Here's a different take on Teen Spirit, by Tori Amos. It's quite slow, and very moody. youtu.be/HaAI3jI7uCc

PM 2Ring

12:26

@Jakobian You might like this song from Television, Friction. youtu.be/vkXDUMQ6nLM They were in the New York scene in the early mid 70s, contemporaries of Patti Smith. Tom Verlaine has a quirky voice, which some people can't stand. ;) But I love his guitar sound, and the riff in Friction is awesome.

robjohn

@CowperKettle there wasn't as much history back then.

CowperKettle

Yes, history tends to accumulate with time.

Humanity should announce an International Year Without History, and do nothing worthy of historical record the whole year.

For the benefit of schoolchildren.

robjohn

@CowperKettle That would be historic

CowperKettle

:)

Jakobian

@CowperKettle does history have isolated points?

shintuku

12:39

yeah

Jakobian

If history were a topological space, how would it look like?

shintuku

some sort of network-like structure

Jakobian

so like a graph

CowperKettle

Oh, I even don't know what "isolated points" are.

shintuku

yeah like a graph

but thing about history is it's not the same as time, it doesn't flow unidirectionally

Jakobian

12:41

$x\in X$ is an isolated point if $\{x\}$ is a neighbourhood of $x$
for example $\{0\}\cup (1, 2)$ has $0$ as an isolated point

it's topology

冥王 Hades

Out of curiosity, when do events become “history”? For example, when will current events be printed in a high school or college textbook?

shintuku

depends, for high schools for instance there are bureaucratic entities that decide the curriculum

冥王 Hades

Isn’t the same true for colleges?

shintuku

it's less centralized for colleges

Jakobian

when governments decide what should we lie about next time

冥王 Hades

12:44

in other words, whitewashing history? It’s surprising and unsurprising at the same time, how most people here are oblivious to the horrific war crimes committed by the imperial Japanese forces

its very rarely mentioned let alone taught

Jakobian

like the biological testing on the Chinese?

冥王 Hades

I’ve been hearing stuff about Florida school boards teaching that slavery was somehow beneficial to African Americans and that being enslaved was better than being dead, makes me want to puke

user 726941

History gets written by the winners of wars.

冥王 Hades

I’d rather die than be a slave to anyone

@Jakobian it gets much worse than that

Jakobian

@冥王Hades I have no context to comment on that

obviously it sounds controversial without context...

冥王 Hades

12:47

@Jakobian it’s pretty blatant once you read it. Here:

nbcnews.com/news/us-news/…

robjohn

why delete "Me & Bobby McGee"?

user 726941

I messed up the punctuation :(

Jakobian

@冥王Hades I don't really get it

CowperKettle

With AI, you can invent your own history.

冥王 Hades

(Hey, since when do unaccredited bodies get to have a say on what is being taught)

robjohn

12:52

@user726941 You can edit comments within 2 minutes.

冥王 Hades

I don’t understand why the U.S. feels such a strong urge to ignore/deny its past. I never understood this

user 726941

@robjohn Thanks 🙏

Xander Henderson

@ThomasFinley TL;DR

You need to edit that question to focus on the key isse.

Soumik Mukherjee

@冥王Hades true, most of the people I know thinks that Japan helped India for its Independence. They have no idea about the atrocious crimes Japanese forces did when they made their way to Andamans

Xander Henderson

In any event, looking at the quoted text, it seems to me that all that the author is saying is that there are two ways of measuring error: relative error and absolute error. The book goes on to say that relative error is generally the more relevant measure of error.

Jakobian

12:59

@冥王Hades my point was that everyone teaches their own version of history that benefits them

CowperKettle

Cool, I never heard of "Andamans" before.

user 726941

Key words are suppressed by the conqueror.

Like replacing "war" with "special operation."

Jakobian

Or like saying that Fermat didn't prove Fermat's last theorem

CowperKettle

An old woman in a clinic queue said that her son, if he receives a callup note, will go to Ukraine, because "if we don't win in Ukraine, Americans will come here and put us in concentration camps".

user 726941

Like they did with the Japanese.

冥王 Hades

13:06

@Jakobian yeah, and I disagree with that approach to history

user 726941

After pearl harbor

冥王 Hades

I don’t see how lying to people does any good

Jakobian

I don't think it matters if you disagree with it or not

CowperKettle

In Russia, Volga Germans during WWII were simply sent to the steppes in Kazakhstan, and many died. There was a lack of housing and facilities and doctors etc.

冥王 Hades

Case in point, I didn’t lie about the fact that I laughed at the waiter who tripped and fell. I could’ve lied and made up a story to make me look like a superhero but I didn’t

Jakobian

13:09

Russia is a bit of an extreme case with all of their propaganda

冥王 Hades

@user726941 it was more than just concentration camps unfortunately

@Jakobian is it because the entire propaganda machinery/structure set up during Cold War was never truly dismantled once USSR went down?

user 726941

We will never know the truth because they lost.

26 mins ago, by user 726941

History gets written by the winners of wars.

Jakobian

it's because the communists came back to power, putting all their friends in the places of power

冥王 Hades

ahem they were Putin all their friends in places of power

Why are there KGB agents outside my home

Jakobian

maybe people don't realize this, but those people are leftovers from the communist era

that's where Putin learned all his tricks from

propaganda mostly affects the elderly and we see it in examples

user 726941

13:16

and the very young

Jakobian

yes, because of their erratic way of thinking

user 726941

~~erratic~~ plastic

Jakobian

but the young have wider access to the news etc.

the real news that is

user 726941

the net has transformed media

you can find any thing you want to believe in

Jakobian

I'm pro-Ukrainians, I mean of course, it was Russia that invaded, using their propaganda as an excuse
Plus who knows what will happen if Russia actually wins...

user 726941

13:20

that much I learned over the pandemic

Jakobian

they're probably committing war crimes as we speak

it's not our war and all.. but I feel like it kind of is our war

sunny

I have a basic question. We can't talk about uniform convergence on a singleton set, right? The reason for the question is the following theorem in these lecture notes.

> Let $$\sum_{n=0}^{\infty} a_n (x-c)^n$$ be a power series. There is a non-negative, extended real number $0 \le R \le \infty$ such that the series converges absolutely for for $0 ≤ |x − c| < R$ and diverges for $|x − c| > R$. Furthermore, if $0 ≤ \rho < R$, then the power series converges uniformly on the interval $|x − c| ≤ \rho$.

I think $\rho$ should be strictly greater than $0$.

Jakobian

@sunny we can

it's just pointwise convergence

$f_n$ is uniformly convergent on $\{x\}$ iff $f_n(x)$ converges

sunny

ok, well then, that sorted that out :) thanks

冥王 Hades

@Jakobian even if it were our war, I don’t know if we’re in a position to fight another full blown war. Don’t get me wrong I’d love to see a real life F-22 vs SU-57 battle but of course that’s a horrible idea

Jakobian

13:26

I mean yeah we don't participate in it, but I think Ukraine needs any support it can get

and not like people just bail out when things get too hot... like Germany

in that sense it should be our war

冥王 Hades

If you’re saying we should send them aircrafts and other stuff, well sure I guess. The problem is Ukraine doesn’t have the necessary military infrastructure to use these systems to their full capacity

And changing that is much harder than just sending them F-16s

CowperKettle

From Wikipedia: ru.wikipedia.org/wiki/…

sunny

14:34

This is a small detail, but it confuses me.

> Theorem. Suppose that $a_n\neq0$ for all sufficiently large $n$ and the limit $$R=\lim _{n\to \infty }\left|\frac{a_n}{a_{n+1}}\right|\tag1$$ exists or diverges to infinity. Then the power series $\sum_{n=0}^\infty a_n (x-c)^n$ has radius of convergence $R$.

> Proof. Let $$r=\lim_{n\to\infty}\left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right|=|x-c|\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|.\tag2$$ By the ratio test, the power series converges if $0\leq r<1$, or $|x-c|<R$, and diverges if $1<r\leq\infty$, or $|x-c|>R$, which proves the result.

I'm confused about factoring out $|x-c|$ in $(2)$. This is only possible if $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|$ exists. The author has specifically written that $\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$ exists or diverges to infinity. In the case of divergence to infinity, $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|$ may not exist either and so $(2)$ may not hold. Is this poor wording or am I missing something?

PM 2Ring

Mar 1, 2022 at 14:38, by PM 2Ring

Here's some psychedelic grunge from some young Ukranian ladies. The Sixsters - На каву

Mar 10, 2022 at 18:22, by PM 2Ring

The Sixsters - Правда (official audio 2022)

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The Sixsters - Психоделіка (official video 2020)

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The Sixsters - For us... For me... (official video 2023)

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Xander Henderson

14:51

@sunny You are being overly pedantic.

The author could write $$\lim_{n\to \infty} |x-c|\left| \frac{a_{n+1}}{a_n}\right|, $$ and note that this either converges (in which case the $|x-c|$ comes out as a factor), or diverges to infinity.

Or, if you prefer, we can work on the extended real number line, with the understanding that $\infty \times x = \infty$ for any $x > 0$.

Jakobian

@sunny If $c\neq x$ then you can do that

Xander Henderson

The statement is, from a pedantic point of view, not justified. But it is one of those elisions in mathematics which makes the exposition easier.

@Jakobian I don't think that is relevant. The confusion seems to be about pulling the factor $|x-c|$ out of the limit.

Jakobian

How is that not relevant

Xander Henderson

@Jakobian Because, again, the confusion seems to be about pulling the $|x-c|$ out of a possibly divergent limit.

Jakobian

and...?

I'm telling sunny that the taking out $|x-c|$ is valid if $x\neq c$

this is very relevant

sunny

14:58

But it's only possible if the limit converges, no?

Xander Henderson

@Jakobian No, because if $|x-c| = 0$, then you done already f'd up by writing $\frac{(x-c)^n}{(x-c)^{n+1}}$.

So at that step of the argument, you already know that $x-c \ne 0$.

Jakobian

now you're the one being pedantic

Xander Henderson

@Jakobian No, the value of $x-c$ simply isn't relevant to whether or not is passes through the limit.

Because it seems rather clear that the question being asked is about pulling a constant of a possibly divergent limit.

Jakobian

@sunny the laws are more general than that, $\lim_{n\to\infty} xa_n = \infty$ when $a_n\to\infty$ and $x > 0$

Xander Henderson

7 mins ago, by Xander Henderson

Or, if you prefer, we can work on the extended real number line, with the understanding that $\infty \times x = \infty$ for any $x > 0$.

sunny

15:01

@Jakobian I see.

Xander Henderson

Exactly.

Jakobian

@XanderHenderson This is implied. I am also pointing out that this is implied by what I said.

Xander Henderson

@Jakobian I'm not disagreeing with you, just pointing out that you repeated what I said.

I am disagreeing about the relevance of $x-c$ possibly being zero, which is completely irrelevant to the question which was asked (about what happens when $|a_{n+1}/a_n|$ diverges.

Jakobian

If you are disagreeing with that then you just didn't understand what I was trying to imply... simply that

Xander Henderson

@Jakobian Ah, so your unclear communication is my fault. Groovy.

Jakobian

15:09

what was unclear about it

Xander Henderson

@Jakobian I have no idea. You told me that my disagreeing with you is a result of me being too stupid to understand what you are saying. So clearly, something you've said is just beyond my ability to comprehend. The failure to understand is entirely on me.

Jakobian

who called you stupid? me? No I didn't

Xander Henderson

6 mins ago, by Jakobian

If you are disagreeing with that then you just didn't understand what I was trying to imply... simply that

You blamed me for not understanding what you have said.

Jakobian

You are equating someone not understanding something with them being too stupid to understand it

not me

Xander Henderson

Don't blame your readers for misunderstanding what you have written.

Take responsibility for your own lack of clarity.

@sunny If you want to be really pedantic, the argument should go something like the following:

If $x = c$, then $\sum a_n (x-c)^n = \sum 0 = 0$. If $x\ne c$ then define $$r = \lim_{n\to \infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n (x-c)^n} \right| = \lim_{n\to\infty} |x-c| \left| \frac{a_{n+1}}{a_n} \right|. $$

By hypothesis, $\lim_{n\to\infty} |a_{n+1}/a_{n}| = R$, where either $R$ is a real number or $R = \infty$. If $R$ is a real number, then $$ r = \lim_{n\to\infty} |x-c|\left| \frac{a_{n+1}}{a_n} \right| = |x-c| \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = R|x-c|. $$ Otherwise, $$r = \lim_{n\to\infty} |x-c| \left|\frac{a_{n+1}}{a_n}\right| = \infty. $$

(Then continue the argument from there.)

But this seems overly pedantic (because (1) in the development of the theory, you have already shown that a power series centered at $c$ converges when $x=c$, and (2) limits involving positive numbers and infinity are sufficiently "well behaved").

Thomas Finley

15:21

0

Q: In a cube of nominal size $5$", the uncertainty in the measurement of each side is $10%.$ Find the uncertainty in the measurement of the volume.

In the calculation of the volume of a cube of nominal size $5$", the uncertainty in the measurement of each side is $10%.$ Find the uncertainty in the measurement of the volume. I tried solving the problem as follows: Given, the nominal size of the cube is $5$ inches. So, if the side length of th...

solution-verification numerical-methods approximation standard-error

Hey guys! Need a little help with this!

sunny

@XanderHenderson Thank you. Just a small edit: $R= \lim_{n\to\infty} |a_{n}/a_{n+1}|$.

Xander Henderson

> In the calculation of the volume of a cube of nominal size 5
", the uncertainty in the measurement of each side is 10

10 what?

What are the units?

Thomas Finley

" stands for inches

I thought this was standard, ig?

robjohn

but 10?

Xander Henderson

There are no units there on the 10.

Jakobian

15:23

@XanderHenderson Please stop with the aggressive behavior toward me

Thomas Finley

@XanderHenderson i am sorry, the latex code for % is not working

Xander Henderson

@Jakobian I have repeatedly pointed out that you are constantly jumping down my throat and contradicting me.

robjohn

@ThomasFinley you need to use $\%$

Xander Henderson

Nearly every time I write anything here, you respond to tell me that I am wrong.

I am getting tired of it.

Thomas Finley

@robjohn oh! Thanks

robjohn

15:25

@ThomasFinley % in latex is the beginning of a comment

so it needs to be escaped

Thomas Finley

@robjohn and @XanderHenderson I have fixed it.

Thanks for pointing that out

Jakobian

@XanderHenderson Why are you trying to make me into some kind of villain? I don't think you are stupid I never called you that. I don't jump on your throat, what even makes you think that. You should calm down

No, I think you are a very smart person actually

I'm just responding to things, I do respect you as a person, please don't take my behavior so personally

Xander Henderson

@Jakobian And once again, you are blaming me. I am really tired of this.

sunny

@XanderHenderson if $\lim_{n\to\infty} |a_{n+1}/a_{n}| = R$ is a real number, i.e. the limit exists finitely, is then $\lim_{n\to\infty} |a_{n+1}/a_{n}| = \lim_{n\to\infty} |a_{n}/a_{n+1}|$?

Xander Henderson

No.

One is the reciprocal of the other.

sunny

15:36

Ok, got it.

Xander Henderson

If $\lim x/y = R$, then $\lim y/x = 1/R$, n'est-ce pas?

sunny

Indeed.

Thomas Finley

I think none of the answers in the option is correct.

Instead, I feel the answer should be, $1/2*(10^{-50})$

What d'ya all think?

Alessandro Codenotti

I don't know what's the issue (and I don't care about knowing), just a reminder that you can block people in this chat if you don't want to see their messages (they can still be seen in the transcript)

robjohn

@PM2Ring: a physicist's answer ;-)

PM 2Ring

15:51

Indeed!

Soumik Mukherjee

How to share the link of a comment?

PM 2Ring

@SoumikMukherjee It's attached to the timestamp of the comment.

Soumik Mukherjee

Oh

PM 2Ring

And if you post a comment link in chat (with no other text), it gets one-boxed. That is, the comment text is displayed. Like this:

The absolute error in the edge length is $0.5$". Since $4.5^3=91.125$ and $5.5^3=166.375$, the maximum absolute error in volume is $41.375$, which is $33.1\%$ of $125$. So I get the same answer. — robjohn ♦ 20 mins ago

@ThomasFinley I think you're right and they're wrong.

OTOH, there is a little ambiguity. "To the 50th decimal digit" could mean to the 50th digit after the decimal point, or it could mean there are 50 sig figs in total, including the digit before the decimal point. But of course that has no bearing on the crazy answer options given in that image.

冥王 Hades

16:10

Good news: the history professor may just curve the grades a bit which could mean I might not have actually failed

I’d hate to retake the exam so that sounds nice

Thomas Finley

@PM2Ring thanks for confirming!

@PM2Ring yes

Thomas Finley

16:34

I need some help in understanding this question.

For some reason, the question looks indecipherable to me.

The question was to determine, whether the statement given is true or false.

I thought the answer to be true apparently.

@PM2Ring Will you mind commenting your opinions on this problem. It would have been really helpful.

PM 2Ring

@ThomasFinley Once again, there isn't enough context. If we carry out all operations exactly, of course there is no truncation error. However, if our numbers are limited to a fixed size, eg 64 bits, then when we multiply two 64 bit numbers the result may require upto 128 bits, so there can be a truncation error.

Thomas Finley

@PM2Ring ok, but I am talking in general. Can we assume that we aren't working in a computer ?

If that's the case, i.e we are working on pen and paper then, I feel that there won't be any truncation error.

Isn't it?

Jakobian

@Thorgott I realized that inside theorem of 1.3.2 of Real-algebraic geometry by Bochnak, Coste, Roy, they actually do prove that this is order preserving. I just assumed they didn't because they state a lemma and a proposition inside the proof of that theorem.

So I would have the full proof if I read it a little bit further

PM 2Ring

16:50

@ThomasFinley That's why I said we need more context. If we're working on pen & paper, certainly there's no truncation. But the given answer only makes sense if we are using a computer (or some other restricted procedure).

@Jakobian Did you like any of the music I linked earlier?

冥王 Hades

17:05

Ted Shifrin

@ThomasFinley But numerical analysis is all about working on a computer!

Soumik Mukherjee

An Indian professor was forced to resign for writing a research paper on possible electoral manipulation. Funny how today is the independence day of India

PM 2Ring

Thomas Finley

17:34

@TedShifrin ha ha...I didn't realize it, tho...I am just a beginner in the topic. I thought it's all about how to work with approximate numbers.

Ted Shifrin

Nope. :)

Thomas Finley

@TedShifrin noted.

冥王 Hades

Numerical analysis classes are a pain sometimes

Transcript for

$Mathematics$ Mathematics