just your definition?

@user58512 Yeah, I am falling asleep here, so I cannot tell whether I was right or not. :) What you said in that message is definitely right.

@AlexanderGruber, $$\int_{0}^{1} \frac{e^{2 \pi i \theta} d\theta}{e^{\pi i \theta}}$$ i think

@peoplepower what was the question?

@AlexanderGruber, what do you think

@Tobias It's proving 3p does not have a primitive root when $p\ne 3$ is prime.

you mean $p > 3$?

Of course.

funny how the question singled out $3$

are the real primitive dirichlet characters exactly the quadratic characters?

well I thought that $$\int_{|z|=1}\frac{1}{\sqrt{z}}dz=\int_{0}^{2\pi}\frac{1}{\sqrt{e^{it}}}ie^{it}‌​dt=\int_{0}^{2\pi}i\sqrt{e^{it}}dt=2\sqrt{e^{it}}\mid_0^{2\pi}=0$$ but I don't think that is correct because (first of all) Mathematica gives me that the integral should be $4i$ and (second of all) $\frac{1}{\sqrt{z}}$ has a singularity at $0$, so the fundamental theorem of calculus shouldn't hold for a contour integral enclosing that

@AlexanderGruber, good pount about the i coming out, but you didnt use sqrt(e^x) = e^(x/2) which I did

it's not a closed contour is it?

yeah, it's around the unit circle

oh, I was thinking about half a circle because of the square root, my mistake

it's weird that the value of the integrand doesnt match up at the end points of the contour

if I just use $\int_{0}^{2\pi}ie^{it/2}dt$ then I get -4 which is closer but I'm still not real sure where the mistake is

right - i think that's the point, that the singularity messes up the FTC

maybe different values come from branch cut choices of where to choose the sqrt

did you try to compute the integral using the residue at z=0 rather than directly?

maybe that gives a different ansewr too

no, i have no idea how to do that. i'm pretty new to complex analysis (and analysis in general actually)

there's a way to compute integrals by just adding all the residues of each singularity

it's related to how an integral whose contour doesn't contain any poles will always be zero

just see the picture here to get the rough idea en.wikipedia.org/wiki/Residue_theorem

hm ok

but the pole has order 1/2 or something, so I don't know how to deal with that

@skullpatrol yes

My only problem is the conversion of $\cos 2 \theta$

@Tobias hey.

hi

@BenjaLim, can i ask you about dirichlet character at some point?

@Tobias Do you know what's going on here?

@user58512 I haven't studied Dirichlet characters, sorry to say.

ok

you say "my question is a little from these two" needs +different

@user58512 Thanks bro :D

Corrected it.

hmm, those do not seem to be isomorphic

@Tobias Guessed there was a problem there.

$A\otimes B/I$ is isomorphic to $(A\otimes B)/AI$ right?

and in your case, $T\otimes R$ is a lot bigger than $T$

wait how did you get the first isomorphism

is it $A \otimes (B/I)$?

yeah

wait what are you tensoring over

@N3buchadnezzar $\cos(2\theta)=\frac{x^2-y^2}{x^2+y^2}$

@BenjaLim let me just check that I recall that correct

hmm, actually I guess it does not make any sense

what does not make any sense...

what I wrote

JT: Not making sense since 1986.

well, I guess it makes sense if you interpret $AI$ as $A\otimes I$

@Tobias wait a minute I'm editing my question on main.

so an obvious map would be sending $a\otimes (b + I) to (a\otimes b) + AI$

..my teacher wrote a lemma with two parts and only proved one.. and I can't prove the first using the same idea

@N3buchadnezzar Sorry, I first gave $\sin(2\theta)$

this sucks

@Tobias see the edit.

hmm

@Tobias so I need that isomorphism badly.

you have a V instead of a W twice in your big quotients

lemme correct it.

@Tobias corrected a few other typos.

I still don't think those are isomorphic. The right hand side is just a polynomial algebra over $R/I$ in $m+n$ indeterminates

hmmm.

but isn't that the same for the left too?

not unless there is something special about $I$

@Tobias It is the ideal of functions that vanish on the given algebraic set.

right, that could be basically anything

well, any radical ideal

yea

@Tobias I should go to bed now.

@Tobias Thanks for the discussion.

well this was a lot of answers quickly: math.stackexchange.com/questions/296375/…

this really annoying

I don't know how to get my notes in order for this class

What's the class?

analytic number theory

Oof. Can't help.

I really need to get organized or ill lose track and have to drop it

Heya

Wow, USPS is finally ending Saturday delivery

Wait I wasn't finished watching that

here i.imgur.com/haaMzft.gif

@EdGorcenski But packages, mail-order medicines, priority and express mail would still get delivered on Saturday~

Yeah

The mail is so slow around here that I don't receive the weekly grocery flyers until the day the deals expire.

So aside from any packages I might get, this seems like a move that won't affect anything.

how am i going to untangle all this

i need to find an example of two groups $H$ and $G$ such that $H$ is not isomorphic to $G$ but $\mathrm{Aut}(G) \approx \mathrm{Aut}(H)$

Wouldn't $\mathbb{Z}_n$ and $U(n)$ work?

what's U(n)?

The set of relatively prime integers less than $n$

hm

with multiplication $\text{mod}\ n$

It is proved that $\text{Aut}(\mathbb{Z}_n) \approx \text{Aut}(U(n))$ work?

and i am pretty sure they are not isomorphic

I don't know what Aut of those groups is

@Eric right, those are not isomorphic

if you remove the Aut from tghe right hand side, they are isomorphic

@Tobias So I am right?

good

wait, they ARE isomorphic?

i am confused?

@Eric no, those two automorphism groups are not in general isomorphic

oh ok

you're saying Aut[Z/(n)] = (Z/(n))^*, that's cool

How does one express that P is any polynomial, where $a x^3$ is not the smallest element ?

@Eric try seeing what you get if you look at the Klein 4 group

i see why that's true now ,jus multiplication by a unit

I found this in GAP

gap> StructureDescription(AutomorphismGroup(SymmetricGroup(3)));
"S3"
gap> StructureDescription(AutomorphismGroup(DihedralGroup(4)));
"S3"

D6 as well has automorphism group S3

right, that was the example I had in mind

In general, most symmetric groups are isomorphic to their automorphism groups

@Tobias, oh im sorry I didn't know D4 was the same as the klien four group

yeah, it is the non-cyclic group of order 4

is it difficult to show S_n = Aut(S_n)?

gap is going into an infinite loop or something when I do StructureDescription(AutomorphismGroup(SymmetricGroup(6))).

@user58512 $n=6$ is an exception

ah "(A6 . C2) : C2"

i have no idea how to read it, but it finally got it

I recall the proof not being completely easy but not using any big machinery

the : means semidirect product, the . means a non-split extension

Can someone help with this terminology: 'reflect a tetrahedron in the midpoint of its altitude'. Does this mean: 'reflect the tetrahedron in the plane parallel to its base passing through the middle of the line between the centre of its base and its tip'? Thanks, it's a bit too trivial to post on SE but has been bothering me a while nonetheless.

@Alyosha I'd say that is the proper interpretation.

hello

hai

@JasonBourne Hello sir

@JasonBourne I'm on isomorphisms now

@Eric isomorphisms of what?

Ok, fair enough.

@Tobias Of groups

ahh, cool

@JasonBourne How are you today?

Wait, what? No, I haven't found any dirt on you (yet).

@JasonBourne I am just asking you. Anyway, I hope your situation gets better.

@JasonBourne Well lets put it this way, the "Introduction to Mathematical Proofs" course is not a prerequisite for number theory, and "Advanced Calculus 1 & 2" are not required for complex analysis, it is a similar situation for Vector analysis too

hey

hi

what's up man?

The roof.

@JasonBourne When I discussed this with a professor, I was told that the only courses that are designed to be rigorous courses are "Advanced Calculus 1 & 2" Abstract Algebra 1 & 2 (2 has not been taught for 7 years) and "Modern Geometry"

@Eric where is this?

@JasonBourne Well it is not a big university, but the have BA and MA type degrees. so it is not a community college

@Tobias The University of North Florida.

@Eric it sounds strange. At university level, all courses ought to be rigorous IMO

@JasonBourne Yeah I will. Definitely.

@Tobias Well, this is in Florida.

and in Florida, math is not done rigorously?

@Tobias 50% of all high school students in the state do not even graduate

well, that does not really have anything to do with what level university courses ought to be at

But then a math degree from those universities will be considered worthless by other universities when it comes to applying for Ph.d positions and similar

Undergraduate math education has been drifting away from rigor for quite some time.

@Tobias In my city the average is about 60%, the public schools are very bad. If you expect a but of high school kids whose highest math ever taken was "Precalculus Algebra" to balze through calculus 1,2, and 3 and dive head first into Rundin in their sophomore year, then your are highly mistaken.

^if you expect a kid

Such a "kid" is rare.

@Tobias Well the professors are great and they will allow you to take graduate level courses, which are essentially what an undergraduate SHOULD take

There is also the issue that low-level undergraduate math courses serve multiple majors: economics, engineering, chemistry, biology, etc.

So there is little teaching of rigor, because a biology major simply won't need it (for better or for worse).

@EdGorcenski Yep.

As such, a math major doesn't encounter rigorous presentation of topics until their upper-level courses. In the last two years, a math student will take only about 10 math courses.

@EdGorcenski Also there are so few student in the math program, that the program is actually losing money

Only about 1/3 of the major elective courses here are ever taught for my school.

I think the bigger issue is that at most US Universities, the Math Department is responsible for covering the math General Education courses

So the vast majority of educational resources of the department are thrown at people who won't ever go beyond Calculus.

@JasonBourne I didn't finish Highschool.

Hey guys, I have a discrete math question

ok

What's the difference between an explicit formula and a recurrence relation?

@Eric Do you have opportunity for inter-university courses?

@EdGorcenski Thank you!

@jtm22, I guess explicit formula is one without recursion

@EdGorcenski Well, I am married, so no. I can't go to another uni

Which means another city here

hmm

@jtm22 An explicit formula gives you a value: f(n) = x. And done.

A recurrence relation is a relation that relies on an earlier value of n

f(n) = f(n-1)+2, for instance.

@JasonBourne You like books. Have you ever read Spivak's "Calculus"?

how would i represent a Factorial but with adding?

as a formula?

as in 12! = 12+11+10+9+...

n(n+1)/2

1+2+...+n = n(n+1)/2

Ed..nice..what about that value in recurrence form?

@JasonBourne My grades in HS were horrible and when i finally realized i wanted to do math, i figured it would be better to just drop out and self study Spivak and get good grades at a community college.

This is a summation, not a factorial.

f(n) = f(n-1)+n

@JasonBourne Which worked out nicely, I got a 3.85 GPA from the Community college and I was allowed to test into advanced calculus 1 in my university.

Ok, great

For recurrence relation I used..6.9 Billion (1.011)^n

@JasonBourne Yep.

is that explicit or recurrence?

anwyay back to work

bye

THat's explicit: for any given $n$, you need not refer to an earlier evalauation.

hmm ok..how would i refer to that as recurrence then ?

because there's just one N

because we don't even need to refer to n-1 at all i believe

Right

If you wanted to make it a recurrence relation, just do f(n) = f(n-1)*(1.011)

But that would be silly.

that is the same thing as raising 1.011^n times 6.9 billion?

In a recurrence relation, you have to look back at the function at some prior step -- or steps -- to determine the value at the current step.

right

Well, if f(1) = 6.9*(1.011)^1

right but that isn't recurrence

that's explicit

Then f(2) = f(1)*1.011 = 6.9*(1.011)^1*(1.011) = 6.9*(1.011)^2

OH

shouldn't we use

and f(c)

g(x)

to differentiate?

I have no idea what you're asking

hello

ok i was just saying you seem to be using f(2) in an instance where f(x) is denoting two different functions

Thanks

f(x) denotes a relation (either explicit, or recurrence). f(2) is a number. f(1) is a number. A recurrence relation just relates a number to prior numbers in a sequence.

Alright, awesome.

THanks man.

Hi guys

hi

hi

?????

I tried to get all the first 10 terms then add them but they doesn't equal to 90 :( ?

-8-6-4-2+0+2+4+6+8+10+12+14+16+18+20 = 90

-8-6-4-2+0+2+4+6+8 = 0

@user58512 mad addition skills ;-)

@user58512 you'r correct

my book says the answer is 15

I mean 15 terms

hi :)

but how'd you do that?

I get 2 answers

n=10 & 9 :(

Sn=n/2[2a+(n-1)d] ?

@user58512 how'd you substitute values in it ?

@devWaleed n(n+1)/2 works for integers from 1 to n

it doesn't work for negative integers, or by skipping integers. However, it's not that hard to modify it

2+4+...+20 = 2*(1+2+...+10) = 2*(10*(10+1)/2) = 10*11 = 110

I know, but do I need to put any value for n ? or just solve it and get n² in end?

-2+-4+...+-8 = -2*(1+2+3+4) = -2*(4*(4+1)/2) = -4*5 = -20

110-20 = 90

n is the number you're adding up to.

my second last step, I get n²-9n-90=0

So if you want 1+2+3+...+100, n = 100

Yes Tn = 100

is it n-1 or n+1 ?

1+2+...+n = n*(n+1)/2

But I have this formula: Sn=n/2[2a+(n-1)d]

why?

we are studying sum of A.P

It means n=10,9 is wrong

you're using n in too many places

Use a different index.

o.0

I used n for a very specific thing: the largest number up to which you are adding.

You are using it to represent the number of terms. These are different things.

The above question was from my book's example.

They are the same thing only when you are adding the arithmetic progression of the integers starting from 1.

I told you exact thing...

I don't know what a or d are

I don't know what your book is asking/presenting.

a= first term

d= a2 - a1

second term - first term = d

ok

So what's your issue?

We are asked to find the number of terms

that make 90

which is 15(answer)

starting from -8?

But I solved and get two answers like the book has one in its example....

yes

If you solved and got two answers, then one is wrong.

actually, both of em are wrong

Or, maybe there are two answers!

Then you're solving for the wrong things.

I get n=10 and n=9

Sn = 90

n[2(-8)+(n-1)2]/2

n[-16+2n-2]/2

n[-18+2n]/2

what should be the next step ?

oops sorry

n(-18+2n)/2 = n(-9+n) = 90

exactly

I did that...

I get n²-9n-90=0

n²-10n-9n-90=0

n(n-10)-9(n-10)-0

(n-10)(n-9)=0

Use the quadratic formula

n=10 , n=9

okay

lemme..

n= 9/2 +/- sqrt(9^2+4*90) = 9/2+/- 21/2 = 30/2, -12/2

Obviously n can't be negative, so you're left with 30/2 = 15.

Your error was in saying n^2-9n-90 = n^2 -10n-9n-90

wait lemme complete quadratic formula

this is not true.

afk for lunch

okay wait

why was my error there?

in breaking?

I get n=15 and -6 now

you added in -10n without accounting for it elsewhere

n^2 -10n -9n -90 = n^2 -19n -90 =/= n^2-9n-90

so, n != " - " therefore n = 15

I was wrong with breaking, ok :|

Thanks for help...

np :)

:(

:\

:|

:/

:)

:S

alt + 253 = ² :)

On Ubuntu

no clue how to do that :P

Windows

^_^

Meh, I prefer Ubuntu

I'm waay faster on it, and I used to use Windows for the majority of my life

I don't know If anybody in my area knows about any other OS

Most common is Xp or Win7