Hey :D :D :D

@Jeff: Hint. $(z/w)^*=z^*/w^*$. (Let's use ^* for conjugation so we don't get it confused with fraction bars.)

@anon ok. I backed up a little, here is where I am now:

No, go back to $z$!

@MattN so 'XXX' means many kisses? Can be, but I thought it is less romantic :D

Argh!!

@BenjaminLim hi, why are you laughing?

no, just smiling

Write $w=e^{it}$ so now you have (I'm ignoring the 1/2 of course) $$\frac{z+w}{z-w}+\left(\frac{z+w}{z-w}\right)^*$$ Now use the fact that $(a/b)^*=a^*/b^*$. Now rationalize.

@JasperLoy Foster's? :)

Remember how you turned a ratio of two complex numbers into (a+bi)/r form, with r real, in basic complex analysis?

yes @anon

you mean by multiplying by the complex conjugate of the denominator divided by the same?

Yes! But you have to rationalize each fraction separately. They'll both end up with the same denominator...

right

i remember

@Ilya ... : )

should i do that now with$\frac{z+w}{z-w}+\left(\frac{z+w}{z-w}\right)^*$?

I should be going now, but given that anon's taken over...

First write the second part as $\frac{(z+w)^*}{(z-w)^*}$, then do it.

See you guys later.

See you tb.

@tb: bye

@tb bye. thx

$$\frac{z+w}{z-w}+\frac{\left(z+w\right)^*}{\left(z-w\right)^*}$$

I think bottom should be w-z?

$\frac{z+w}{w-z}+\frac{\left(z+w\right)^*}{\left(w-z\right)^*}$

@tb Bye!

No, it should be z-w.

Well, wait, I might have reversed $z,w$ somewhere.

@BenjaminLim Hi, can I have a question to you?

Whatever, just keep pushing forward :P

ok. i think $(w-z)^*$. i'll work from here to figure out the processs and then do it correctly on the homework

so i think i need the $(z-w)^*$.

$$\frac{z+w}{z-w}\frac{(z-w)^*}{(z-w)^*}+\frac{(z+w)^*}{(z-w)^*}\frac{z-w}{z-w}$$ $$=\frac{|z|^2+wz^*-w^*z-|w|^2}{|z-w|^2}+\frac{|z|^2+w^*z-wz^*-|w|^2}{|z-w|^2}=2\frac{|z|^2-|w|^2}{|z-w|^2} $$ set $z=e^{it}$ or something

I forgot: can we reopen this? Looks like a perfectly legitimate meta question...

Bye for real now.

ok. give me a minute to look at that

hey @anon

hey

would you mind looking over an argument?

Sure, but again, I'm probably no use for your stuff ;)

i want to show that if $M$ is a f.g. $A$ module (A comm), $N$ another $A$-module, $S$ a multiplicative subset, then $$S^{-1}\text{Hom}_A(M,N)\simeq \text{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}N).$$

@anon still working

Bwahaha. It would take me 5 minutes just to wrap my head around what that says.

i think i have an idea for a proof, you might understand it if i tell you

i'm not asking why it's true, after all

unless you've never seen this kind of thing at all

No, I'm squirting my brain acid on it for digestion.

$(w+z)^*=w^*+z^*$??

I can do that, after all.

@Jeff: Indeed.

i'll just talk to myself and hope someone listens. so i want to define a morphism from the lhs to the right $\Phi$, by $\Phi(f(a_1m_1+\cdots+a_n m_n)/s)\mapsto f(a_1m_1+\cdots+a_n m_n)/f(s)$. this is a morphism and it's well-defined

I take it $S$ is closed under inverses then.

Why not just replace instances of $S^{-1}$ with $S$ then?

yes, that is what is usually meant by a multiplicative subset of a ring

err

that is what is meant by $S^{-1}$

you are inverting a multiplicative subset

you are MAKING it closed under inverses

$(z-w)(z^*-w^*)=zz^*-zw^*-z^*w+ww^*=|z|^2-zw^*-z^*w+|w|^2$

you are localizing the ring at $S$, letting you make some fractions

@Jeff: Don't bother splitting $(z-w)^*=z^*-w^*$ by linearity, just apply $aa^*=|a|^2$ to $a=z-w$.

I don't know of localization, the best I can do is think of it as a group of fractions, but whatevs

(instead of "field of fractions")

@anon, ok

that's exactly what it is @anon, it's just a group

but you are localzing the ring by letting a multiplicatively closed subset be invertible

so if you take the integers, localizing at (2) means that you now allow fractions $m/2^n$

and this doesn't change the isomorphism type of the ring

and how come $w^*z-wz^*=0$?

Yes I see. It's like $\bigcup_n 2^{-n}\Bbb Z\subset\mathrm{Frac}(\Bbb Z)$

@Jeff: It doesn't. Notice how things cancel when you combine the two fractions.

yes

@anon Oh! :D. OK, i understand the last equation you posted

Anyway you seem to have the pertinent isomorphism, and just need to think about the technical check-to-make-sure-this-works stuff.

this: chat.stackexchange.com/transcript/message/4468704#4468704

yeah it's just the surjectivity that's bothering me

(are you helping two at once! omg)

injectivity and well-definedness are clear

@ymar yes?

Is the backwards map well-defined? If so, then it's surjective right? :P

@BenjaminLim According to this the ring $\mathbb R[x,x^{-1}]$ is a PID, right?

@Jeff More like 1.2 at a time, as yours I can do without thinking and Eric's I barely even understand.

@anon lol

these test questions are so ridiculous

how do I get the derivative for $5^{xcosx}$

@BenjaminLim $\mathbb R[x]\subsetneq \mathbb R[x,x^{-1}]\subsetneq \mathbb R(x)$

@Jordan: Hint $$a^{f(x)}= e^{f(x)\log a}, $$ now just use chain rule

and I am suppose to have that memorized for a test? ridiculous

The formula I wrote? Pretty much, yeah.

f(x) = cosh(3x+5) find the derivative ,then there are ones for inverse cosh, inverse sin, how is this testing what I have learned in the class? It is just testing what I have memorized

So, at this point I've got $\displaystyle Re(\frac{e^{it}+z}{e^{it}-z})=\frac{|w|^2-|z|^2}{|w-z|^2}$

Oh look JM dropped in.

and I need ...

or what does asking a word problem involving a cone test me on? It isn't even taught in the class

That LHS should be $\mathrm{Re}\left(\frac{w+z}{w-z}\right)$

@anon Well...

$\displaystyle Re(\frac{e^{it}+z}{e^{it}-z})=\frac{1-|z|^2}{|e^{it}-z|^2}$

bbl

^ Jeff: Yes.

I think I need to take calculus a third time

@anon yes $w=e^{it}$

how is a problem like $y = (sinx)^{lnx}$ even possible without memorizing obscure rules that are incredibly rarely used

wait! no

@Jordan Third time's the charm. The third time I took my fourth semester of analysis, I passed.

that big R means "Real"

I can't afford a third time

or the humiliation and shame involved with it

the first time I dropped it I just made up the excuse that I wanted a higher grade

One more time:

@Jordan Come on, swallow your pride. Is dropping out less humiliating?

@anon is that right?

@ymar I would rather accept that "math isn't what I want to do" or some "it isn't my calling" bullshit then just accept that I am well below average intelligence

that means that $|w|^2-|z|^2=1-|z|^2$?

No. The "Need" equation should replace $w$ with $e^{it}$ consistently throughout the equation. That's just a matter of specializing $w=e^{it}$ (notice that $|e^{it}|^2=1^1=1$.)

oh wait! that's true, isn't it? because $w=e^{it}$ has radius 1

@Jordan Which year is that?

what do you mean year?

thanks anon. i think i have it now. i'll replace the need appropriately, but i have the work down.

@Jordan of your studies. I mean how many years you admit to have wasted when you decide to drop out.

Umm, I know what "noch wach" means @MattN.

Thank you @tb, I was AFK.

And Matt.

@ymar I am not sure, I have over 100 credits

@Jordan I have no idea what this means :)

Like 3 years of school I guess

So you're close to your bachelor's title?

no, like 3 more years

but at this rate I have to take every math class 2 or 3 times it means about 12 more years

huh? six years for a BA?

yeah or seven

Wow. So how long does it take people to earn a Master's?

6 years?

So you get both titles at the same time?? I don't understand this.

I didn't take any college classes in high school so I had to start with pre algebra, college algebra, college algebra, trig, calc 1, calc 1. that is 3 years right there

no I am just stupid so I take 6 years, normal people take 4 years

also intro to physics, college physics, physics 1 and physics 2, language 1, language 2, intro to chem, chem 1, and all my generals

that is about 3 years

then I can start my degree

or major

OK, this system seems to be completely different from the one we've got in Poland.

@Jordan It could take four years (best) to six here as well, it doesn't have anything to do with stupidity.

it is probably similar

@Jordan Well, there's no such thing as college classes in high school here.

I think the system is set up like that so colleges can get more money out of people, college costs a lot here. Between $10,000-$30,000 for cheap in state schools

I consider physics, chemistry and math beyond geometry to be college classes

I don't know, I feel your pain. I'm also considering dropping out but not because sucking at math hurts my pride. I just feel I'm wasting time.

Juhuu, I got a badge.

Ooh, a shiny one too.

I just feel like there is too much to learn in a math class and there is never any review. I spend all my time learning the new material (but not enough time to learn proofs or anything, just memorize the conclusion of the proof) that I don't really have time to cover log rules and other stuff

Show: $\displaystyle \frac {1}{2\pi} \int_a^b K Re(\frac{e^{it}+z}{e^{it}-z})dt = K\frac{b-a}{2\pi}+\frac K\pi \arg(\frac{1-ze^{-ib}}{1-ze^{-ia}})$ (hint: switch the order of integration). There's only one integration? What order?

@jordan in math class, everyone learns at the same speed! you just upon my pet peeve about math teaching.

everyone pays at the same speed

@Jeff Perhaps the hint means to say: switch the order of Re and $\int$.

How can I do that. The Real part contains the variable being integrated over

I just don't understand how people keep these rules of math internalized after not using them for months or years

that is what I can't do, I can learn the rule fine and I am great at memorizing but I can't remember them unless I use them

I memorized the summation formulas for i i^2 and i^3 but I will forget them unless I use them

@Jeff: If you want a conceptual explanation, think in terms of the Riemann sums, and remember that Re() is continuous so it interchanges with limits. Specifically, $$\lim\sum_i \mathrm{Re}\big(f(t_i)\big)\Delta t_i = \mathrm{Re}\left(\lim\sum_i f(t_i)\Delta t_i\right).$$

This works because the differential dt is real.

Also remember that $\mathrm{Re}$ is $\Bbb R$-linear..

oh. so integrate first, then find the real of that.

but we can get rid of the $\mathrm{Re}$ using chat.stackexchange.com/transcript/message/4468704#4468704

(what we just did)

@Jeff $\int (u+iv)\mathrm{d}x=\int u\mathrm{d}x+i\int v\mathrm{d}x$

and I use "we" advisedly.

Just wait on that man. Do the inside first.

Oh, duh. I should have said that.

are you telling me to wait on using what we did, or wait on doing the integral (i always confuse myself in chat room convos by asking questions too fast)

Wait on applying the Re() operator.

@Jeff :-)

@robjohn you've noticed, have you? :D

@Jeff I remember having to say "wait a minute" once or twice :-)

ok. so then back to this: how do I integrate $\displaystyle \frac {1}{2\pi} \int_a^b K Re(\frac{e^{it}+z}{e^{it}-z})dt$? I can't split it into $u$ and $v$ because it's alll Real part.

@robjohn well, if you're going to volunteer to help me again, you could keep that in your clipboard and just paste it in.

@robjohn or we could just work out some easy to type code, like let "w1" mean SLOW DOWN, JACKASS! (short for wait 1 min) :D

@Jeff: Notice what happens if you apply Re to what robjohn wrote, replace u+iv on the LHS with f(z), and u on the RHS with Re(f).

Well, probably better to use x instead of z. You get $Re \int f dx = \int Re(f)dx$

Isn't this true:

@Gigili Not at all...

Perhaps you're confusing it with $\sin x/x$ as $x\to0$.

Why not?

@anon Oops, I'm sorry.

As $x\to\infty$, $|\sin x/x|\le 1/x\to0$.

@Gigili $\displaystyle\lim_{n\to\infty}\frac{\sin(n)}{n}=0$

Got it, thanks.

@Gigili Since $|\sin(n)|\le1$ and $n\to\infty$

I wonder if there's a way to find the most common name in math.SE. There certainly are a fine amount of Davids, Jonases, and Alexes...

I love sharp estimates. Bounding $\sin$ by one is not really sharp 8-).

@JM But there is only one real Jonas, like there is only one real Batman...

@JonasTeuwen what would you bound it by?

@robjohn $1$ in this case 8-)). Or an odd number of terms of its power series! 8-).

Say 101 terms. That's a nice upper bound.

@anon thanks. we just don't have time to figure that out before class time and abandoned that part of it.

@JM I have a question about a question! Today I was teaching complex analysis and I asked them to derive the series for the Bessel functions from the generating function. I wonder if it would be an appropriate question here if I asked for the physical interpretation of the generating function ($\exp(\frac12 (z - 1/z) w)$). I know that for Hermite polynomials it has to do with some random walk which makes much sense (Ornstein-Uhlenbeck operator!)

@anon OK. I notice what happens. I forget if we said what we were going to do first, the integral or switching the real and integral.

@JonasTeuwen Ooh, mixing things up! Usually people derive the series from the DE...

@JM Yes.

Anyway: sounds like a nice question. (That I don't know the answer to.)

But this is a course in complex analysis 8-)).

Not in differential equations!

Okay, thank you!

So, you applied Cauchy's formula to the GF, I take it...

Out of context that sounds very funny.

@Jeff are $a$ and $b$ specified?

@Jeff: Sorry mate you'll have to nab someone else to help you like robjohn. I need to go get 2 days of sleep.

@rob not sure what you mean or what expression you're referring to

(^ GF=girlfriend)

@anon It does, doesn't it...

@anon np. thanks so much for your help (also, class starts in 20 minutes, so you were free anyway!) :D

@anon Since GFs are by definition, not singular, they integrate to 0 :-)

Like taking the Fourier transform of one's cat.

Anyway zzz...

@Jeff I'm sorry, can you click on the arrow next to my comment to see the referenced message?

@robjohn i click the arrow, but i don't see a reference to what you were replying to.

@robjohn we're out of time, anyway. thanks for helping.

@JM Here it is! math.stackexchange.com/questions/140958

@JM Yes.

@JM And no. I also just used the binomial theorem.

@Jeff sorry for being late.

@robjohn you weren't late - they waited until yesterday to do a semester project.

after they go to class i have an free 30 minutes i will look at it

8-)!

@JonasTeuwen You might want to add special-functions also. :)

$\int_p(yz dx + zx dy +xy dz) = \int_p \nabla (xyz)\cdot d\bar{r}$, path integral right?

$d\bar{r}=i dr + j dr + k dr$?

$\nabla (xyz) = (\partial_x + \partial_y + \partial_z) (xyz)=yz+xz+xy$

now there is a mistake

I need some unit function there.

yes!

$\nabla (xyz) = (i \partial_x + j \partial_y + k \partial_z) (xyz)=i yz+j xz+k xy$

Now it makes sense

Wha__?

$\int_p(yz dx + zx dy +xy dz) = \int_p \nabla (xyz)\cdot d\bar{r}\\=\int_p (i \partial_x + j \partial_y + k \partial_z) (xyz)\cdot d\bar{r}=\int_p (i yz+j xz+k xy)\cdot d\bar{r}$

@hhh Isn't it simpler just to note that $\mathrm{d}x=i\cdot\mathrm{d}\bar{r}$, etc.?

where $d\bar{r}=i dr + j dr + k dr$, now I think it is of the right idea? (I mean dot product works as intended etc)

@JM Sure!

@robjohn Sorry can you elaborate why?

(i cannot see your point directly)

$\mathrm{d}\bar{r}=\mathrm{d}(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})$

@robjohn Yes! That is clever idea, good thinking!

why is Pi used, what is pi?

I am trying to find the concept "käyräintegraali työintegraali" in English. The words mean something like curve -integral and work -integral. Ideas?

Example problem:

$\int_p \nabla u \cdot d\bar{r} = \int_p \nabla (xyz)\cdot d\bar{r}$ from $(-1,1,-1)$ to $(1,2,3)$.

@robjohn: Do you still have that rendering problem?

I need to hit the bookmark each time a $\LaTeX $ message appear.

@N3buchadnezzar It should converge to 3.54744407881674811090936660730

@Gigili I did last I looked

@robjohn I noticed that as well, perhaps i wrote it incorrectly.

@Gigili me too.

@Gigili yes, I do

but it is not that bad, helps debugging...

@Gigili Oh, that rendering problem. That is a change in the chat code I think.

@N3buchadnezzar Ah, that converges to 3.14159268251190473490351169088

Which is a decent approximation to pi

@robjohn Might be it, Chat rooms seems to have problems recently.

Also pi = 355/113 and e = 163^(32/163) = (pi^4 + pi^5)^1/6

runs away

@N3buchadnezzar You've been reading that xkcd comic, I take it...

@JM Alas, none of those approximations are in that comic.

math!

@Jordan Correct.

Let's see... play videogames, or do math? That is the question.

@AntonioVargas Depends. Which one's more fun?

math is never fun, so that is easy

On different days I'd have different answers :)

That's pretty much expected. :) What about today?

I think I failed calc 1 the first time because of Skyrim, Diablo 3 beta and Battlefield 3, probably put about 600 hours into those games in a semester

Good lord. I have about enough time for ~20 hours of gaming per semester!

@Jordan Now that you admitted it.

Maybe if I stopped visiting reddit and stackexchange I could bump that up 3x...

I cant focus on math for long so I hae to do a problem then do reddit for 5 minutes

Alright, I'm going to go install KOTOR.

@AntonioVargas I or II ?

Well, I'm off for today. See you kids later...

@JM Have fun.

KOTOR is good, wish bioware could do a story that well again

logarithms are incredibly difficult to keep track of, it is like an upside down way of things, but viewed in a mirror

@Jordan Arent there just like three rules you have to keep track of ?

a lot

then derivatives, antideratives

power trickery, ln tricks, log tricks and all sorts of weirdness

$\log_a(a)=1$ , $\log(ab)=\log(a+b)$ and $\log(a^b) = b \log(a)$

This should be all.

I dont even remember what log means

what number to a will give me a? it is 1

never heard of the a + b one, how does that work?

log_a(b) = c is equivalent to a^c = b

so if I am trying to find the number that will make a certain number that other number I just add the two numbers and they will as a power make the original number the result?

@Jordan If you insist everything is a trick or weird, your mind won't really accept it.

@Jordan Something like that $\log(5 \cdot 5) = \log (5) + \log (5) = 2 \log (5) = \log (5^2) = \log (25)$

that seems counter intuitive

logs are just really hard to picture

Since logarithms are very closely related to exponents, and exponents work that way. Why should not logarithms do that?

it is like an algebra problem set up backwards

so log base 4 1 is = 0?

$ e^a \cdot e^b = e^{a+b} $ and so forth

it is just hard to imagine how these are manipulated

$ \log_a{1} = 0$

what is e?

If anyone can give me a hand here, it will be much appreciated.

@Jordan Some arbitary number

ln is log base e?

Fine, I just used e without much thought. Here let me state it even more explicit.

Let c be some arbitary real number, and same with a and b, then

$c^a \cdot c^b = c^{a + b}$

@PeterTamaroff You body diagrams seems right to me

so if I am trying to find a number raised to a power that is raised to a power I can just find the power of the first number times that power?

Well, I'm reading the rest of it.

@Jordan huh?

Uh-oh, it isn't too difficult as you said!

@Jordan Too much power in there.

@Gigili :4472383 Yes, I'm OK with that, the problem is the motion of the system.

log is just like short hand for an algebraic expression to find $a^x = n$

@Jordan Indeed

But it can be used and applied to much, much more.

I dont understand how hte log tricks apply to the algebra expression

They are not tricks...

$10^x = n$ is the same as $8^x + 2^x = n$

@Jordan NO!

Is it true that $(\Bbb{Z}/2\Bbb{Z})^{\times}\times(\Bbb{Z}/4\Bbb{Z})^{\times}\cong(\Bbb{Z}/8 \Bbb{ Z } )^{\times}$ ?

Sorry for the rudeness.

Take the case x=2

@Jordan If you are not comfortable with symbols exchange the symbols with numbers, then you see that your example fails.

100 = 68?

but my example is the log trick

@Jordan What log trick?

log (ab) = log a + log b?

@Jordan Jordan, they aren't tricks, they are identities.

so $4^x + 2^x = n$ is the same as $8^x = n$

@Gigili Are you looking at the problem?¿

that doesnt seem true

@Jordan Umm no.

@Jordan Jordan, what you want is $4^x \times 2^x = 8^x$

is you take logs, taht becomes

$\log 4^x +\log 2^x = \log 8^x$

but the trick is log ab = log a + log b

@Jordan IT IS NOT A TRICK! Please.

and if log is just a^x = n then why doesnt that work?

@PeterTamaroff Yes, aren't they both $T$?

identitity, whatever

@Gigili Sorry? I assume the magnitude $t$ to be the same for all vectors $\vec T$, $\vec T_0$ since the rope is ideal-

@Jordan Trying to deep dive, when you do not know how to swim is not wise.

@PeterTamaroff I thought you wrote $T$ and $T'$

See this

Never mind.

@Gigili Oh, well. Those are the reactions to the others.

@N3buchadnezzar Now that is really clever.

@PeterTamaroff Well, I don't know what's the second equation in your system

@Gigili It is the equation of the other block. There the tension pulls up (in the negative direction) and weight pulls down (on the positive direction).

@PeterTamaroff I know, but the other block is $m_2$

@Gigili Their masses are equal.

Oh, well. Let me finish reading then.

I think you have to include $N$ somewhere, it doesn't make sense otherwise.

Also, I hate Physics.

N-mg \sin = 0

@Gigili Agg, I hate this type of physics.

Then N=mg \sin

@Gigili I'll add that

@Gigili sin of what? 37 I presume

this is why logs are so confusing, when you apply logic to the rules they don't hold

@Jordan When you apply logic to your statement it doesn't hold.

@Jordan My logic works fine with logarithms, have you considered getting a new set of logic? =)

@PeterTamaroff You should have both 37 and 53, no?