math.stackexchange.com/questions/129460/… makes me sad

@anon, that is a weird question...

I am very intrigued by «So far, my testing has not broken anything.»

what was tested?

Did coffee still taste good after modifying the value of $\zeta(0)$?!

This probability question has stabbed me in the guts until now. But, now I think I know how to solve it.

Hi @MarianoSuárezAlvarez Good to see you drop by every now and then!

@Mariano I wanted to bring to your notice this link...

Of course, the reward is pittance but still...

@AnnaLear Please anonymize this question: math.stackexchange.com/questions/112173

@Zev May I ask why? Was that a request by the user or some such thing?

BTW, a warm welcome to this room. :)

@KannappanSampath That's correct, it was a user request. And thanks for the welcome!

@ZevChonoles Will do.

@AnnaLear Thanks!

@ZevChonoles And done. :)

Does this mean the upvote by this user on some of the answers to this question will be taken away? @AnnaLear

@KannappanSampath No, the votes aren't affected at all. We simply remove the account association between the post and the user. (Anyone can request that, by the way. All posts made on SE are licensed under Creative Commons, which also stipulates that while you grant us the right to reuse the content forever, you can get your name removed from it if you don't want to have it attributed to you anymore.)

@AnnaLear I see. Thank you for the information

Anytime. :)

pfft! you mean if i write a book on here, you've already got the publishing rights? well, i never!

Yes, but, I am curious why would someone do that here, other than Arturo, who of course has already done that...:)

well, obviously, Arturo is an attention streetwalker

@DavidWheeler Hey thanks for the heads up on a subgroup acting on the dodecahedron

@DavidWheeler It is NOT trivial to see that $M \otimes_A N \otimes_A k \cong 0 \implies (M \otimes_A k) \otimes_k (N \otimes_A k) \cong 0$

@BenjaminLim np

hey @KannappanSampath

Hi @Ben.

@KannappanSampath I understood that thing I was having trouble with.

It turns out it was some problem in my understanding of base extension

That sounds great. It is actually so much fun, you know...:)

So, now time to move on to next concept in AM?

@KannappanSampath your next concept?

For me it's the next problem

For me it is $\S2.4$ Direct sum and product of modules.

But, not until this month would end. I have an array of exams approaching.

Ok. Concentrate on your exams first.

All my final exams for this semester exam have been scheduled b/n 2 and 9th May.

@BenjaminLim Sure.

I suggest you drop commutative algebra now and concentrate on those exams

Like 2 weeks before my field theory and analysis exams, I dropped everything else and concentrated on those

Precisely what I am doing.

@KannappanSampath Because trust me when you want to do tensor product you WILL need to drop everything else to concentrate on it.

@KannappanSampath If you need any help with Tensor Products give me a ping I will try to help you

@BenjaminLim Yes. That's like the ultimate goal for this chapter. So, I want to do it slowly.

@BenjaminLim Sure thing. I'd do that.

@KannappanSampath Exact Sequences as well. Make sure you really get your head around that. It is useful for understanding why say you have that $A/a \otimes_A M \cong M/aM$

Like in last chapter---last few sections were new to me.

oh good, you can explain them to me, then....

what is "a"?

Man the mean value theorem is so hard to understand

@DavidWheeler an ideal I'd think.

@DavidWheeler Ok first do you understand why the sequence $0 \rightarrow a \rightarrow A \rightarrow A/a \rightarrow 0 $ is exact?

Won't that follow from the definition of the canonical quotient map?

@KannappanSampath What will follow?

Kernel for $A \to A/a$ is the image of the map $a \to A$.

Badly said, but you get that, no?

@KannappanSampath Oh you meant the exactness of the sequence not the result about tensoring

yes that follows

@Jordan Say your average speed in 30 minutes is 15km/hr

@BenjaminLim Bah, I told you I don't know tensoring, so all I can tell you is about exactness!

then the mean value theorem is telling you that at least once in that half an hour your speed was indeed 15km.hr

if a is an ideal sure, it's FIT for modules

or rings, whatever

@DavidWheeler So in general the tensor product does not preserve exactness

but it is right exact

So tensoring with $M$ gives us the exact sequence

define "right exact"

$a \otimes_A M \rightarrow A \otimes_A M \rightarrow A/a \otimes_A M \rightarrow 0$

@DavidWheeler When tensoring with an $A$ - module $M$ gives that the sequence above is exact

exact, or right exact?

well the sequence above is exact because it is exact at each term

But I should not be saying this because I don't know it:

The tensor product is an example of a contravariant functor $F$

So if $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is zero

we say that $F$ is right exact if $F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow 0$ is exact

@DavidWheeler does this help?

how would I factor something like $2-2x-x^3$ I dont want a walkthrough I just dont know how to do that, do I have to use a computer?

what happened to the 0 in front?

@Jordan...look for rational roots first

I dont know what that means

try dividing by x -1, x - 2, x + 1 and x + 2 first, before you seek other options

also I guess I meant find the zeroes/roots of not factor

finding roots and factoring are essentially the same

It goes away

I am not sure how to divide by x-1

an easier way than dividing straight away would be plugging in -2, -1, 1, 2, no?

because if you have a zero in the front

and if it's zero then you can divide

@Tyler same thing.....i always test the roots first, but if i do find one i'll have to divide

it would mean that the map $a \otimes_A M \rightarrow A \otimes_A M$ is injective

which is not always true

Yeah I can't factor this

ok, so tensoring does not preserve the injectivity we could have $r_1 \otimes m = r_2 \otimes m$?

@DavidWheeler Well yes, but recall that elements in the tensor product are not always elementary tensors like the above

@Jordan there is a lemma that tells you that if you can factor over the rationals, then you can factor over the integers

they can be sums?

@ Benjamin I have no idea how to factor this though, like the process used to get an answer out of this. There doesn't appear to be any way to manipulate this problem to isolate an x

@DavidWheeler Exactly. In general an element in $M \otimes N$ is a linear combination of elements of the form $(m_i \otimes n_i)$ with coefficients in the ring

this is why math is so frustrating, the very first problem, the easiest one and I can't do the first step

@Jordan Suck it up man. Do you want to handle tensor products? I don't mean to be harsh

But let me tell you it can't be factored

Well I don't know what to do for this problem at all if i can't even do the first step. I have absolutely no idea how to factor this and I cant find any guides on the internet

so the $m \otimes n$ are like a basis

@DavidWheeler Well if you want to think about it like that it's ok

But let me warn you in general there is no "basis" for the tensor product

@Jordan It has no rational root, so by Gauss' Lemma can't be factored over the integers

i didn't mean to imply "like a free module" more like the ideal product IJ

I don't know what that means, wolfram is telling me there is a root at .7709 which my calculator agrees with

@Jordan Let me explain it clearly to you

@Jordan, why are you trying to find a root?

First what is your level? Undergrad?

to graph it

@DavidWheeler That is right.

Jordan is attending community college

The fact that an element in the tensor product is a linear combination of elementary tensors comes from the fact that we quotient out the free module on the product set $M \times N$ by the submodule generated by those usual relations that give us bilinearity

It crosses the x axis, it has to have a root

This is like my third year of college, but yeah undergrad

@Jordan Can you tell me why if we can factor a cubic over the rationals, this means that it must have a rational root.

Why????

Tell me.

I dont know what factor a cubic over means

@Jordan first plot (0,f(0)), and then plot (-1,f(-1)) and (1,f(1)), to get a few points established

then find the critical points, so you know when the slope changes direction

I am just following the guide in the book that says to start with intercepts, then other stuff

@Jordan You want to factor over the integers, rationals, over the reals or over the complex numbers?

because if you want to factor over the complex numbers your polynomial splits completely

finally, take the second deriviative to look for inflection points

It has to contain the splitting field of your polynomial

@BenjaminLim he's not trying to find a root, per se, he just needs to graph it so he knows what it looks like

I thought he was trying to factor the polynomial?

@Jordan Well if you know Cardano's method it is easy.

no, he just wanted to find a root (x-intercept)

and already there is no x^2 term so this should be easy

intermediate value theorem?

cardano's method is NOT easy...it's a horrendous mess

?

I just want the x intercept, and I have 2-2x-x^3

@Jordan Do you know the intermediate value theorem?

Yes

ok, Jordan, the y-intercept is easy, right?

There is a calculus way of finding the intercepts!

Yes the y is easy, it is 2

Let $f(x) = 2-2x-x^3$

$f(0) = 2$

$f(1) = -1$

and the $x^3$ term is negative, so it starts out high on the left, and low on the right?

@Jordan So by the intermediate value theorem, what now?

by continuity, we know it crosses the x-axis somewhere between 0 and 1

@DavidWheeler Well the $x^2$ term is not there

@BenjaminLim yes, i see that

So therefore the IVT tells you that there is a root somewhere.

the derivative is equal to $\frac{f(0) - f(1)}{0-1}$

the image of a connected set is always connected under a continuous function

so it will be closer to a pure cubic like $-x^3$

@DavidWheeler That makes it easy!

no, Jordan, the derivative is $-2 - 3x^2$

@Jordan The derivative of 2 is zero

the derivative of -2x is -2

Hey @Ben While making sense, you are actually making it harder for Jordan. I'd suggest we keep quiet and leave David to handle the scene. :)

@KannappanSampath Why are derivatives involved when he wants to find the roots?

since the derivative is always negative, it is montonically decreasing

@KannappanSampath Actually the polynomial has no rational roots by the rational root theorem.

@BenjaminLim He is actually graphing, so, he would be looking at concavity, convexity, maxima and minima and all of that

however, it DOES have an inflection point: take the 2nd derivative

I mean I am just trying to find intercepts first, if I don't find the intercepts my graph will be wrong

@KannappanSampath oh he wants to graph it? I had the impression he was trying to factor it. That was what he said no?

which is -6x, which is 0 at x = 0

I am trying to factor it with the intent of finding the zeroes so I can graph it

so we have an inflection point at 0

@Jordan well you won't be able to find a simple factorisation of it

this means it is shaped like a sideways "lazy S"

@Jordan In fact if you try to factor it into two polynomials with rational coefficients it is not even possible

I dont understand why such a hard question is even in this book, it isn't teaching me anything about calculus or graphing

@BenjaminLim He said that initially because finding a root for that was hard for him. Now, David is telling him he does not have to know that precisely yet have a decent graph.

ok

On the test I am going to just cheat and plot out each point by hand on my calculator but it would be nice to be able to do the other stuff I am suppose to know how to do

to get a little more detail, it is concave upwards left of the y-axis, and concave-downwards to the right of the y-axis

@Jordan Some general advice: Anything that tests your patience, that makes you frustrated, that is your TEACHER.

if you plot (-2,f(-2)) and (2,f(2)) as well (5 points so far), you should have a fairly decent picture of what it looks like

I am just suppose to know intercepts, even or odd and asymptotes memorized

it has no vertical or horizontal asymptotes, but it does asymptotically approach the cubic $-x^3$

@Jordan do you understand what I wrote above?

so for REALLY BIG x, it looks almost identical to $-x^3$

Yes but I hate getting frustrated, I can't ever learn like that

But, unfortunately, life is not so good.

@BenjaminLim i don't think Jordan has ever studied "just polynomials", he's probably only seen them as functions to graph, or solve problems of optimization with

man

the good news, Jordan, is that polynomials are continuous functions, so if you know what it's doing at a point, it behaves "pretty much the same" as long as you don't go too far away from that point

I dont understand this problem at all, there is no zero for the derivative, but the second derivative has a zero

that's because it has no local maximum OR minimum

but how do I know I just wasn't good enough at algebra to be able to factor it properly?

it "swoops down" from negative infinity pretty fast, and starts to flatten out (but still going down) near the y-axis, crossing it with a slope of -2.

between 0 and 1, it starts to take a nose dive again, crossing the x-axis as it does so

these problems seem way easier if I ignore calculus and just calculate all the points

well, at the point (1,-1) it has a slope of -5, that's about -78 degrees (fairly steep, right?)

it just seems like the point of that problem was to show me how little algebra I know

saying «but how do I know if I just did not see how to factor it?» transmits exactly the same information as «but how do I know I just wasn't good enough at algebra to be able to factor it properly?» without being overly dramatic about this incompetence of yours about which you love to tell us...

I am not complaining, it is a real problem I have on tests

at the point (0,2) it has a slope of -2, which is around -63 degrees, still not all that flat

I don't know if the problem is not solveable or if I am just doing it wrong and I spend excessive amounts of time on it

everyone has that problem—not everyone expresses it as dramatically

well, Jordan, factoring polynomials is not a trivial task....even the pros find it hard sometimes

very few things students do get on my nerves as much as their constant self-deprecation

well most people probably don't have a problem passing calculus

most people don't TAKE calculus....

and most people who do take it have big problems with calculus

and even among the people who do, not everyone passes the course

BTW, why should Calculus be taught at Undergrad at all...That should be done while you are still in high school--Or so I thought.

you see, there's ideas in calculus that arithmetic does not prepare you for

Teaching some real analysis sounds good...

no but I am probably going to fail it a second time, this is like my 6th college math class and I havent gotten above a c yet

@KannappanSampath school curriculums vary...in most american high schools calculus is an "honors" class, not required for graduation

more drama

I see.

@MarianoSuárezAlvarez ? Not understanding what that refers to? :)

I am just saying that I know I am not good at math, it is pretty apparent by now

but we do not CARE that you are not good

we can help you

we can provide guifdance

but we do not CARE that you are bad at it

@Jordan what you ARE good at, is beating yourself up. you should stop that.

that is why all your talking about how bad you are is absolutely irrelevant and off-topic here

saying "i suck at math" repeatedly will not improve your grade point average....and may actually hinder your attempts to improve. positive thinking, dude. use it, it works.

@Jordan If you keep whining that you are not good at it, YOU WILL NEVER BE GOOD AT IT.

What you think, you become.

Well that is who I am, I can't help it but it is a real problem that I have on tests. Each test I take I have to figure out what fractional exponents mean, how to multiply and divide exponents and other basic stuff. I spend a lot of time doing that and then I need to spend further time on other basic stuff to make sure I am right. my biggest problem in math is failing the tests. I can understand most of the concepts I just cant pass tests

@Jordan My first calculus/linear algebra cookbook class exam I got 66%. Guess what I did? I sucked it up!

Why did you take this course then?

There is much more to me than what is apparent from what I say in this room

you do not have to pour yourself into it, you know

simply use it to ask math questions

i understand...it takes you too much time. part of that, is fear, i suspect. you freeze up, and then start going in circles.

@MarianoSuárezAlvarez no, you are the one-dimensional being you appear to be. stop pretending otherwise.

it is just hard to be postitive after failing a class and then taking it again and getting a C only because the other teacher was easier. I never improve at math I just sort of stay just as bad at it but they let me keep taking more

one-dimension... I wish

but I will shut up about it

Hausdorff dimension was made to measure me

@Jordan We can help you. Stop the complaining. Is it going to help you? Answer that.

@MarianoSuárezAlvarez Are you the zero space?

I dont know if it will help, but I probably have to reevaluate myself and possibly consider a different career

I'm $\log_32$ -dimensional :(

@Jordan...your negative attitude puts people off...it comes across as if you don't really desire to learn. if that is not actually the case, then you're not doing yourself any favors.

I want to learn, it just feels like it never pays off

@Jordan, it can often take decades to fully realize the benefits of what you learn. it's a long feedback-loop

I just meant grade wise

@Jordan You are always thinking about grades?

it is all that matters

@Jordan What shallow thinking.

@Jordan, what i can tell you is: if you stop thinking about the grades, and start thinking about the course material, your grades will improve without incessantly worrying over them

well my GPA is to the point where I cant really transfer anywhere

because before I was not worried about grades, so I did not mind failing a class to take it over, but that killed my GPA and now I have pretty much no chance of ever going to any university

...

@Jordan oh stop it stop this nonsense about transferring elsewhere. Why don't you sit down and start focusing on the material at hand? Stop wasting your precious energy and start focusing on course material

@MarianoSuárezAlvarez What do you mean?

someone is giving you bad advice then. lots of universities have remedial programs for people who needs to get their grades up

well not any near me

we've had this Jordanian drama for a few days now

@Jordan Do you have skype?

Yes, I am a witness. Not a pleasant thing to say. :/

jordan...it's a big world...cast a wider net

well I want to go to a decent school, otherwise why even try? If I fail at everything else I can just go to a bad school and continue to fail there, I want to imrpove

IT IS NOT THE SCHOOL THAT YOU GO TO. IT IS WHAT YOU DO WITH YOURSELF. IT IS WHETHER YOU UNDERSTAND THE MATERIAL OR NOT. DO YOU UNDERSTAND THIS????????????????????????????????????

let me put it this way: in the real world, what matters is your ability to do the job

@KannappanSampath skype

a fancy diploma from a first-rate school may open some doors, but if you don't know what you're doing, you'll be back out the door soon enough

well then why go to a "better" school, is the education not better?

the internet is a veritable store-house of FREE STUFF.....free math books, free online tutoring sites, calculational aids, and instructional videos

@MarianoSuárezAlvarez as a moderator please do something about this.

I can kick him from the room

lol awesome

the "best" education is one you actually RECEIVE, even a brilliant teacher doesn't do your learning for you

it's been five days of me telling you to stop whining precisely so that I don't have to kick you

@BenjaminLim oh, you said a naughty word....MOM!!!!!

@DavidWheeler In australia if you want to really emphasize something you use it.

i don't frigging care what the damnation you do in that maggot-infested cornucopia of filth. cease! and desist!

(did i lay that on a bit too thick?)

@DavidWheeler who was that aimed at?

i was being facetious

I Know :D we make racist insults at each other among by friends :D

it's an interesting linguistic phenomenon: close friends often use insults as terms of endearment

cursing is a very interestinbg phenomenon

in Argentina, we curse with multipliers :)

it is difficult to explain in English, though!

is there any easy way to factor $x^2 - 2x +3$?

Does the quadratic formula count as easy?

The roots are complex, so it isn't going to be obvious. You could also complete the square.

I dont really know the quadratic formula

It's worth learning.

To continue the last thought: $x^2 - 2x + 3 = (x - 1)^2 + 2$.

How do you know that?

You could set this equal to zero, or write it as $(x - 1)^2 - (i\sqrt{2})^2$ and factor that, since it's of the form $a^2 - b^2 = (a - b)(a + b)$.

Google "completing the square".

I know the difference of squares formula

@DylanMoreland It is not trivial to see that $ M \otimes_A N \otimes_A k \cong (M \otimes_A k) \otimes_k (N \otimes_A k)$

Well, that was out of the blue.

@Jordan The pointer to Google was more for recognizing this as a good thing to do.

@DylanMoreland AM exercise 2.3 they have $(M \otimes_A N)_k = 0 \implies $M_k \otimes_k N_k = 0$

Was busting my nuts last night to understand that isomorphism

@BenjaminLim is there any connection between A and k?

@DavidWheeler $A$ here is a local ring with maximal ideal $a$. $k = A/a$

I remember the problem. $A$ is a local ring and $k$ is the residue field.

@DylanMoreland the implication follows from the isomorphism I stated above. But it was not at all trivial to see that isomorphism

The hardest part of math for me is remembering all the stuff I learned before, I learned long division a long time ago but I have no idea how to do it now. Does everyone else just always know this stuff or do you also have to relearn it all when it is reintroduced later into math

@Jordan it's hard for me to describe how i internalize stuff. i try to remember "general methods" that can recover specific information if i need it.

for example, i don't recall too many trig identities, i just re-derive them

how do you derive them?

It seems incredibly complicated

using the basic definitions and the fundamental identity $\cos^2\theta + \sin^2\theta = 1$

which is really just the pythagorean theorem in disguise

Dividing by $\cos^2$ gives one and $\sin^2$ gives another.

I dont even remember what sin or cos are I just know that I can punch in the value on my calculator

SOHCAHTOA, lol

I never learned those words, just x, y and r

the various trig functions are (in one definition) just ratios of two sides of a right triangle

> A year passed. Autumn changed into Winter. Winter changed into Spring. Spring changed back into Winter. And Spring gave Summer and Autumn a miss and went straight on into Winter.

A somali guy was my trig teacher, they probably dont use those words

@Jordan I don't mean to be offensive but looks to me like your basics are very weak.

I know they are but I am trying to work on it

^^^ that's what's going on right now.

@BenjaminLim yes, they are....he really should be studying analytic geometry....calculus would be LOTS easier for him, then

@Jordan Like you don't know the quadratic formula, we should be taught in precalc BEFORE taking calculus.

And things like SOHCAHTOA

Man that was like I don't even remember now when I learned it...

I know of the quadratic formula I just dont know why it works

Dylan's method of completing the square is a superior approach

I didnt really take any math until I was in college

otherwise it's just an "arcane formula"

@Jordan it works like this; here is an easy quadratic to solve; $x^2 = 0$

unfortunately, they aren't ALL that simple, but the next best thing is $x^2 = a$

if we can "do a substitution" say u = x + something

@tb @tb was that for me?

we can sometimes turn our quadratic into (x+something)^2 = a

It seems to me that the quadratic formula is just completing the square.

which we can solve, by taking a square root

that's the "strategy" behind "completing the square", to get something easy to solve

@BenjaminLim No. I'm just complaining about the weather and why on earth it had to snow again and why the temperature is what it is right now.

for example, if we have $x^2 + 2a + b = 0$

Utterly useless. We alread had enough days below zero this year.

@tb there is some use to a certain number of days below zero?

the $x^2 + 2a$ part "almost" looks like part of the "sqaure" $x^2 + 2a + a^2 = (x + a)^2$

@robjohn I don't think so. It makes Matt feel good, apparently, and it makes me feel bad. The net result is neutral.

so we cheat, and add in the "missing part"

@DavidWheeler we don't cheat

that gives us $x^2 + 2a + a^2 + b = a^2$

put the "b" on the other side:

$x^2 + 2ax + a^2 = a^2 - b$

now we have a perfect (completed) square on the left

$(x+a)^2 = a^2 - b$

You're missing $x$'s @David

take square roots, and then subtract the a, we're done

why so i am, i need a bigger buffer

i can't fix the older ones, too late

No problem. Just that Jordan is aware did I make that comment.

the point is, the "quadratic formula" is like the "clean-up at the end"....how you get there is MUCH more important

in general, we don't have the "2" present in $x^2 + 2ax + b = 0$

we only get something like $x^2 + ax + b = 0$, so we have to cheat again, and call a , 2 halves of a.

The "classical" formula is for $ax^2+bx+c=0$ which gives $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ but David will get there soon, sorry.

it looks a bit messier, but it's the same thing; $x^2 + ax + b = x^2 + 2(a/2)x + b$

Can we please have a few close votes here?

now the "missing part" is $(a/2)^2 = a^2/4$

@tb above my paygrade ;-)

@robjohn, true enough....but if a is non-zero, we can divide everything by it first, so that's a minor concern

@DavidWheeler I know, I was just relating the formula that is well-known. I will be quiet now :-)

so if $x^2 + ax + b = 0$, then $x^2 + 2(a/2)x + a^2/4 = a^2/4 - b$

now we have "completed the square" because we have a prefect square on the left:

$(x + a/2)^2 = a^2/4 - b$

and again, to finish up, we take the square root, and then subtract the a/2

@robjohn let me draw an analogy: 1) how do I integrate a polynomial? 2) how do I prove the identity $\sqrt{2\pi} = \int e^{-x^2}$ ? 3) please explain the functional equation of the zeta function to me.

there is so much to know with algebra

@tb) how do i factor the difference of two squares? 2) what's a normal subgroup? 3) what is the character table of E8?

for example when I do the newton's method problems for my homework and I am required to estimate the roots of a function to 8 decimal points. How do I know how many roots there are? Is there a simple way to find out the number of roots since the polynomial will be too complex to factor

@David: that would work, too :)

@Jordan...in general, no there's not a quick way to find out how many roots. but...the maximum number of roots is limited by the degree of the polynomial

newtons method problems are horrible, was just hoping there was an easier way to do them

also: odd polynomials always have at LEAST one real root

newton's method is numerical...thank god for calculators

also...newton's method doesn't always work....or to put it slightly more accurately, you need to make a good first guess

facstaff.unca.edu/mcmcclur/mathematicaGraphics/Newton/…

the convergence of newton's method to a root can be less than desirable...

for certain initial guesses, convergence takes a really long time