Mathematics - 2016-12-02 (page 9 of 11)

Null

12:00

my math prof explained it to me really great i think: even tho Q is dense. The chance to have a period in the decimal representation of a number is smaller, the longer the period will be. By that reasoning Q being dense doesn't imply that it has the same cardinality as R. obv not rigorous, but it helped me to understand

Fargle

@Null We could take a simpler function, like $f(x) = \frac{1}{\sqrt{2}}$ when $x \geq 0$ and the opposite for $x < 0$. But even with that, any closely-approximating continuous function (which can be found) cannot satisfy the above relation, as it would have to take every value in the interval $\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$, and thus can't solve the equation you gave.

Balarka Sen

@Brody Depends on whom you ask :)

Null

can someone say what he does to the first $\varepsilon$ in the bottom line? is that a 1, since n>1 or what?

. youtu.be/Pb8HDtZiGYc?t=7m50s

or a n which got hit by a truck?^^

Fargle

@Null It's an $n$ that got hit by a truck.

Null

actually that is the way i tried. But i was really unsure

$(1+\varepsilon)^n>n$ i had so far. i thought you always try to have $<\varepsilon$ at the end

Secret

12:17

[Abstract algebra] Caption:

imgur.com/a/EaW6T

How to impose associativity by careful picking of columns and rows

Sanity check: The 8 possible nxn tables that spawn basically consists of two sets, a n plicate of columns and a n plicate of rows of the middle element being chosen. In other words, one level of Light's associativity test

Therefore theoretically, repeat this n more times you get all $n^3$ associative rules

Heonjin Ha

Simple question I have!

Alessandro

hi @Balarka

is it true that every topological manifold (locally euclidean, second-countable, Hausdorff and connected) can be embedded into $\mathbb{R}^n$ for some $n$?

Heonjin Ha

How can I write the following statement in mathematical statement? :
Let $x=(x_n)$, $n=0,1,...$. any word $w$ occurs in $x$ with bounded gaps for any word $w$ occurring in $x$.

Note that word $w=w_{1}w_{2}...w_{n}$

Balarka Sen

@Alessandro That's true.

Alessandro

one doesn't actually need connected there I'd guess then

I also guess the proof is nontrivial?

Balarka Sen

12:25

I can't write down a proof without further assumptions on the manifold :)

Alessandro

I see

Null

is (a) a short version for saying the sequence $a_n$?

Balarka Sen

@Alessandro Relevant keywords are "partition of unity". I think I can only prove it for compact topological manifolds with those.

Alessandro

that's a name I've never heard before

ok, I've just seen the definition, I don't know how or why they're useful though

Balarka Sen

It patches together local functions into global ones, is all. If you locally embed in R^n (trivial), you can patch them up to get an embedding in R^N (N most of the time is much larger than n)

@Alessandro It's a technical result the proof of which you shouldn't worry about. With some technical assumptions on the manifold this $n$ can be improved to $2\dim M + 1$ (in fact 2\dim M$).

This is known as the Whitney embedding theorem

"Technical" is a bad choice of word. I should say "niceness". But it's not imposing too much conditions in the sense that most manifolds you'd see satisfy them.

Alessandro

12:44

Hm, I see, it was just a curiosity since all of the examples of topological manifolds we've seen are obviously embeddable into R^n

Balarka Sen

yup

@Alessandro Here's a question to ponder on. What are the 2-dimensional manifolds upto homeomorphism? Is there a simple classification?

maybe you should think about that for 1 dimension first but that's easy to guess (not quite that easy to answer though)

Null

i dont grasp how $\binom{x}{2}=\frac{n(n-1)}{2}$. I know it is right, I just seem to be unable to reaarange the terms

ah lol

2!=2 ...

Alessandro

We already saw the answer in class for the 1 dimensional case @Balarka, you get either the circle of R

Balarka Sen

mhm

Alessandro

I'll think about 2 dimensional manifolfs later, I'll have some time to kill on the train

Balarka Sen

12:56

OK

Akiva Weinberger

Are there any special facts about 17 other than constructible heptadecagons and wallpaper patterns?

Null

it's prime?

. en.wikipedia.org/wiki/17_(number) there's actually a wiki article about that number lol

wouldn't thought so

Akiva Weinberger

Apparently 17 is the minimum number of givens necessary for a sudoku puzzle to have a unique solution. Huh.

Null

13:13

@AkivaWeinberger that is quite a special fact

Heonjin Ha

Have a question!

If I search with a tagged one, then it is not in the list.

I mean when.

3

Q: Need help in proving a set has bounded gaps.

Let $(X,B,\mu)$ be a probability space and $T\colon X\rightarrow X$ is measure preserving. Let $A\in B$ such that $\mu(A)>0$. Then I am asked to prove the following claims: The set of positive integers $n$ such that $\mu(T^{-n}(A)\cap A)>0$ is infinite. The above set of positive integers $n...

this question is here and there is a tag 'ergodic-theory'

When I search with 'ergodic', it is not in the list.

How can I search with the tag?

NaCl

@Null hello

Null

13:35

is $(1+a)^x$ much easier to expand than $(a+b)^x$. a,b, in R. x in N

Sophie

$(a+b)^x=b^x(\frac a b +1)^x$

Null

clever

Sophie

I wonder if many mathematicians other than Erdos use amphetamines

Noein

Hello o/
I'm looking for some introductory material on some concepts and I'm not sure if that much literature even exists that discusses them using the terms I'm searching for.
It's confusing enough that I'm no longer sure what I'm searching for.

Null

what is the most uninteresting interesting number?

Noein

13:47

For a start (I'm trying to read the new paper on the quasi-polynomial time graph isomorphism algorithm):
String iso/automorphism
The Coset Intersection problem

Akiva Weinberger

@Null 51?

Null

is it possible that a general case is easier to prove than an explicit case?

Semiclassical

hi chat

Secret

[Abstract algebra] Just figured out in any semigroup why associativity is such a pain to keep track on. This is because the number of possibilities for each step of some product xy in the cayley table being specified depends on what is being done previously. That is, suppose the set of all possible ways to specify the product of an algebraic structue form a space of sort, then the degrees of freedom on what the next randomly chosen element to be specified is path dependent in this "space"

Semiclassical

@Null math.stackexchange.com/questions/1007256/…

Secret

13:55

in The h Bar, 15 mins ago, by Secret

Caption: Checking associativity on the way go, without leaving the cayley table:

NB The example in the illustration is a very general monoid and nothing else being specified

GFauxPas

If I have a parameterized curve in C or R2, and it's closed, how do I find at which point the parameter has the curve start drawing over itself?

Is this a question worthy of posting?

Ali Caglayan

@GFauxPas Do you have a more concrete example of a curve?

GFauxPas

x(t) = Re(t^(4+1i)*exp(-(4-3i)*t)), y(t) = Im(t^(4+1i)*exp(-(4-3i)*t))

It appears to overlap itself when I graph it using R, but I don't even know for sure that's true

I don't know how to check :(

Ali Caglayan

I guess one way of trying would be to equate the equation at t and t + p for some unknown period p

GFauxPas

Should I ask a question?

on the site?

I'm not sure of what effort to show here

I can explicitly solve for Re and Im if that helps

in this case it's not that hard, because it's t^4t^i exp(-4t)exp(3it)

Ali Caglayan

14:10

Well keep it in C to be simple

We postulate that at $t$ and $t+p$ the value of $z(t)=t^{4+i} e^{(3i-4)t}$ will be the same

GFauxPas

right

Ali Caglayan

so plug in t and t+p and equate

GFauxPas

it might not be periodic, the "period" might change, does that make a difference?

Ali Caglayan

I never said it was periodic

Just some value in the future it will be the same

Null

@Semiclassical thanks

GFauxPas

14:13

I can equate imaginary and real parts seperately, right? thanks for the guidance

Ali Caglayan

yes

GFauxPas

nice, thanks Ali

Ali Caglayan

np

Null

. smbc-comics.com/comic/doctor can someone explain this one?

GFauxPas

Ithink he's showing off how much sex he has

Mahmoud

14:35

Hi

DHMO

hi @Mahmoud

Mahmoud

If $r$ is a rational number, such that $r=\frac pq$ and there is no common factor between $p$ and $q$ What do we say about the relation between $p$ and $q$ ?

Hi @DHMO

DHMO

@Mahmoud they are co-prime

Mahmoud

@DHMO Oh, yes .. I never get the point of some Mathematical terms .-.

@DHMO Thanks $:)$

DHMO

You are welcome

Roh

14:41

Hi guys

I have a problem

How can I calculate exp(-26569332.53)

?

Just tried google but it gave me 0

DHMO

@Roh What do you need it for?

Roh

@DHMO To calculate Gibbs free energy.

Gibbs free energy

In thermodynamics, the Gibbs free energy (IUPAC recommended name: Gibbs energy or Gibbs function; also known as free enthalpy to distinguish it from Helmholtz free energy) is a thermodynamic potential that can be used to calculate the maximum or reversible work that may be performed by a thermodynamic system at a constant temperature and pressure (isothermal, isobaric). Just as in mechanics, where the decrease in potential energy is defined as maximum useful work that can be performed, similarly different potentials have different meanings. The decrease in Gibbs free energy (kJ in SI units) is...

DHMO

@Roh Anyway, exp(-26569332.53) is a very small number.... you can't imagine how small it is

maybe if you could give us the whole expression

Roh

Ok, hold on

...

I'm reading a book about combustion

I have its solution manual

DHMO

and where is exp(-26569332.53)?

Roh

14:50

This picture I taken from the solution manual

@DHMO I calculated everything within that parentheses

DHMO

@Roh how does that give you -26569332.53?

@Roh that gives me -8.4861575466

anyway, I have to go now

Roh

@DHMO I just used my "SHARP" calculator

Ops! sorry! i toke photo from the wrong place.

Going to take another photo

Null

in 20 years it's like this: how do i prove this? nevermind, used my $\neg$Brain-calculator

Aksel'sRose

Having difficulty with the following: $\alpha_1=\frac 12 (\alpha+\alpha^*)$ and $\alpha_2=\frac {1}{2i} (\alpha-\alpha^*)$. I need to prove that $\alpha$ is normal if and only iff $\alpha_1\alpha_2=\alpha_2\alpha_1$. Any ideas on setup?

im thrown off by the $\alpha$ being normal when it is inside the given alphas

Null

how do i stop mathjax on one site?

Aksel'sRose

15:14

@Null sorry, I have no idea...

GFauxPas

cos(log t + 3t) = sin(log t + 3t) can only be solved numerically, right?

arctic tern

well, you can solve cos(x)=sin(x) first, then log(t)+3t=x using lambert W

actually evaluating lambert W is essentially numerical, but then again when a calculator evaluates any elementary trigonometric or transcendental function that's also essentially numerical in nature

GFauxPas

hmm, well, assuming that there are solutions, I end up integrating upon a path that traces over itself

contour integral

interesting, I'm not sure what happens then

oh, weird, then each time it comes back to its original point it vanishes?

assuming no poles?

arctic tern

@Aksel'sRose have you tried multiplying out $\alpha_1\alpha_2$ and $\alpha_2\alpha_1$?

Aksel'sRose

I plan to do that for the other direction since its an if and only if. But, I should probably try that first

Steve

15:43

Given an inhomogeneous complex differential equation of this form:
$z''(t)+a\cdot z'(t)+b \cdot z(t)=4\cdot exp(i\cdot t)$
A particular solution to this differential equation is:
$z_0=(4+6\cdot i)exp(i\cdot t)$
Now we observe a real differential equation with the same coefficients as above:
$x''(t)+a\cdot x'(t)+x(t)=4\cdot cos(t)$
Now determine a particular solution of this of the form:
$x_0(t)=c_1\cdot cos(t)+c_2\cdot sin(t)$
Determine $c_1$ and $c_2$. Any hints to how do I solve this?

Kasmir Khaan

16:12

Hi guys i need help

Random Variable

16:31

@DanielFischer I have an question sort of related to what I asked you yesterday. I can show that $\int_{0}^{\infty} \frac{e^{-a\sqrt{1+x^{2}}}}{\sqrt{1+x^{2}}}J_{1}(bx) = \frac{e^{-a - e^{-\sqrt{a^{2}+b^{2}}}}}{b} $ for $a, b>0$. I want to argue that the formula holds for $\text{Re}(a) \ge 0$. The right side of the equation is an analytic function for $\text{Re}(a) >0.$

And the left side is an analytic function for $\text{Re}(a) >0$ since $|e^{-a\sqrt{1+x^{2}}}J_{1}(bx)|$ is an integrable function for $\text{Re}(a) >0$. But how can I show that it holds for $\text{Re}(a) =0$ as well? Does the stuff about Laplace transforms apply here?

Hold a sec. That second part is wrong. The dominating function needs to be independent of $a$.

GFauxPas

Is there a way to see the mathjax rendered here?

Random Variable

@GFauxPas math.ucla.edu/~robjohn/math/mathjax.html

GFauxPas

thanks

Null

16:47

$n+\sum_{k=0}^{n-1}k=\sum_{k=0}^{n}k$?

Lozansky

Let $n$ be a positive integer. The equation defining the $n$th root of a positive number $A$ is $x^n-A=0$. Show that, for an approximate $n$th root with small forward error, the backward error is approximately $nA^{(n-1)/n}$ times the forward error

So I see that $nA^{(n-1)/n}$ is the derivative of $f(x) = x^n-A$ at $x = r = A^{1/n}$

Semiclassical

Hi chat

Astyx

Hi @Semi

How are you ?

Lozansky

But how do I proceed? :(

Secret

Abstract algebra: Improved algorithm that allow construction of an associative algebra on the fly. This also showed how much restraint associativity imposes on the algebraic structure (in this case a magma)

Aksel'sRose

16:56

Having difficulty with the following still: $\alpha_1=\frac 12 (\alpha+\alpha^*)$ and $\alpha_2=\frac {1}{2i} (\alpha-\alpha^*)$. I need to prove that $\alpha$ is normal if and only iff $\alpha_1\alpha_2=\alpha_2\alpha_1$. Any ideas?

user2154420

Question: why does any extension of $\mathbb{Q}$ containing a root of $x^6+3$ have to contain a primitive 6th root of unity?

Secret

Note Sx and xS is the same notion as a coset. Since cosets are equivalence class and they often have "geoemtric properties" in the form of orbits that partition the group structure. This suggests associativity itself in its most abstract form might have a geometric interpretation

Null

@Secret i find your stuff cool

Aksel'sRose

17:14

how about: what is the definition of normal for transformations? or how would you show normal?

Mahmoud

Hi

Aksel'sRose

hello

Mahmoud

Guys is it that $\forall x \in \mathbb R : \lfloor x\rfloor +1= \lceil x \rceil$ ?

Akiva Weinberger

No, let $x$ be an integer

arctic tern

@Aksel'sRose normal means $AA^{\ast}=A^{\ast}A$

Akiva Weinberger

17:21

When $x$ is an integer, $\lfloor x\rfloor=\lceil x\rceil$, @Mahmoud.

Mahmoud

Yes. @AkivaWeinberger

Aksel'sRose

@arctictern I thought i had seen that but it is nowhere in my text! Thank you!

Sylent Nyte

guys, what is the definition of an even integer in terms of n? so, how can you write n to be an even number

arctic tern

have you tried googling "even number"?

gonna have to say what you mean by "in terms of n"

Mahmoud

So $0=1$, :/ @AkivaWeinberger

Sylent Nyte

17:23

@arctictern yes I have, but I get different results

arctic tern

what do you mean?

Mahmoud

But what if we say that it hold for $\mathbb R - \mathbb Z$ ? @AkivaWeinberger

Sylent Nyte

so for example, theres one stating that n = 2k for integer k and that will yeild every even number

arctic tern

yes

Sylent Nyte

then another stating that n - 2k is an even integer

arctic tern

17:24

huh?

Sylent Nyte

i dont really know to be honest, but how else could you write it other than 2k?

arctic tern

I am skeptical that you read "n-2k is an even integer" as a definition of an even integer anywhere

Sylent Nyte

"Assume
n
is an even integer.

Then
n
= 2
k
for some
k
2
Z
, by the de nition of an even integer."

arctic tern

@SylentNyte why do you need a second way to write it? I mean, you could write (2k+1)-1, or anything else equivalent to 2k

Akiva Weinberger

@Mahmoud Yes

Sylent Nyte

17:26

sorry, broken format, straight copy and paste

arctic tern

@SylentNyte you know you can edit things you've copied and pasted before pressing enter? :P

you can also edit and remove comments within a time window

Akiva Weinberger

Two minutes I think

Mahmoud

@AkivaWeinberger Thank you $:)$

arctic tern

you also responded to something I didn't say. I did not say I was skeptical you saw a definition of even integer as "expressible as n=2k for some k." I said I was skeptical you saw a definition of even integer as "n-2k is an even integer" (your words, not mine)

Sylent Nyte

@arctictern ah, well I apologise then in my case

let me rephrase my initial question

i had an exam yesterday, I had gotten the test back today and I need to respond on it. $f(n) = n^5 - 20n^3 + 64n$, prove that for even $n$, $f(n)$ is divisible by $2^5$

so in order to prove this I had written that let $n = 2k$ and hence substituted and solved it from there

then it asked me to prove that for even n, $f(n)$ is divisible by $2^7$

arctic tern

17:32

@user2154420 if $\omega$ is a primitive cube root of unity, then $-\omega$ is the $6$th root. can you express $\omega$ in terms of $\sqrt[6]{-3}$?

Sylent Nyte

I had said let $n = 4k = 2(2k)$ which is even however this does not cover every possible even number

arctic tern

@SylentNyte why did you say let n=4k?

you seemed to understand perfectly well in part (a) that even integers are all of the form n=2k. that applies just the same in part (b). why throw away your understanding or pretend you don't understand?

Sylent Nyte

a wait

shouldve said let $n = 2(2k+1)$

@arctictern because am dumb

Ted Shifrin

Morning, tern.

arctic tern

morn

Mahmoud

17:44

Hi @TedShifrin $:)$

Ted Shifrin

Hi @Mahmoud

Kasmir Khaan

Hello ted

I have a question for you

you helped me before

Ted Shifrin

Sure, @Kasmir.

Kasmir Khaan

@TedShifrin am trying to solve this question , find surface area of z=4-x^2-y^2 , 0<z<1

but when i drew the function z

it lies below z=1 and 0

what can they mean ?

Steve

can anyone help me with my problem?

Ted Shifrin

17:51

It's a downward pointing paraboloid, highest point at 4, going down to $-\infty$. They want only the portion with $0\le z\le 1$.

How are you parametrizing the surface?

Kasmir Khaan

(s,t, 4-s^2-t^2)

Ted Shifrin

OK, but what region in the $xy$-plane does this lie over?

Kasmir Khaan

my guess is a circel of radius 2

Ted Shifrin

When you have circular symmetry (the $x^2+y^2$ is a hint), you should always think of polar coordinates. Or set up the integral in $xy$-coordinates and then switch to polar coordinates.

Kasmir Khaan

but in general ihave hard time finding that

Ted Shifrin

17:53

No, $z=0$ gives you a circle of radius $2$, but what about $z=1$?

Kasmir Khaan

hmm i honestly dont understadn how to tackle this problems

Steve

The problem is:

Given an inhomogeneous complex differential equation of this form:
$z''(t)+a\cdot z'(t)+b \cdot z(t)=4\cdot \exp(i\cdot t)$
A particular solution to this differential equation is:
$z_0=(4+6\cdot i)\exp(i\cdot t)$
Now we observe a real differential equation with the same coefficients as above:
$x''(t)+a\cdot x'(t)+x(t)=4\cdot \cos(t)$
Now determine a particular solution of this of the form:
$x_0(t)=c_1\cdot \cos(t)+c_2\cdot \sin(t)$
Determine $c_1$ and $c_2$. Any hints to how do I solve this?

Kasmir Khaan

what is it am calculating

Sylent Nyte

@arctictern I do apologise but i dont understand what i am doing

Ted Shifrin

@Kasmir: You want the surface area, i.e., $\iint_S 1\,dS$ or $d\sigma$. I don't know what notation your course uses.

Kasmir Khaan

17:54

@TedShifrin does surface area mean the full "body"of the thing

yes that notation is fine we use Y for the region but thats fine

Ted Shifrin

It means you're adding up all the areas of little pieces by doing the surface integral of the function $1$. ... But go back to the region. When $z=1$, what does that give you in the $xy$-plane?

Kasmir Khaan

hmm when z=1 you asking about the projection on the plane paralell to xy ?

it should give a circle of different radius

Ted Shifrin

Right. What radius?

Kasmir Khaan

sqrt (3)

Ted Shifrin

Right. So the region that parametrizes it is $\sqrt 3\le \sqrt{s^2+t^2} \le 2$.

What's your formula for $dS$?

Kasmir Khaan

17:59

its line integral over the region of cross product

but that is not what i dont understand , the set up of the problem

Ted Shifrin

not line integral, just integral

magnitude of the cross product

You have to figure out how to set up a double-integral over a region in your parameter space, just like you did with plain double- and triple-integrals a month ago.

Kasmir Khaan

well it was very easy when it was about double and tripple

but here i dotn get the theory behind this

i mean what shape do i have ?

Ted Shifrin

Once you parametrize the surface, you're back to doing a plain old double integral. You just need to figure out what to integrate and what the region is.

So here we have the region between the circle of radius $\sqrt3$ and the circle of radius $2$.

Kasmir Khaan

hmm okay please give me a second

am trying to work it out

and thanks alot again ! :)

Semiclassical

hi chat

Ted Shifrin

18:06

Sure, @Kasmir. Hi @Semiclassic.

@Steve: You understand that $\exp(it) = e^{it} = \cos t + i \sin t$?

Steve

@TedShifrin Yes

Null

Let $a\in\mathbb{R}^{+},n\in\mathbb{N}_2$.

To show: $(1+a)^n\geq\binom{n}{2}a^2$

\underline{Basecase}

$n=2$

$$(1+a)^2\geq\binom{2}{2}a^2$$
$$\iff a^2+2a+1\geq a^2$$
\underline{Aussumption}

For some $n\in\mathbb{N}_2$ the following holds
$$(1+a)^n\geq\binom{n}{2}a^2$$
\underline{Claim}

Then the following holds too
$$(1+a)^{n+1}\geq\binom{n+1}{2}a^2$$
\underline{Step}
$$\binom{n+1}{2}a^2=\frac{n(n+1)}{2}a^2=\frac{n(n-1)}{2}a^2\cdot \frac{n(n+1)}{n(n-1)}=\binom{n}{2}a^2\cdot \frac{n(n+1)}{n(n-1)}$$

can someone look over this?

Semiclassical

@null don't spam.

Null

at the end i show that the limit of the fraction is 1 right?

@Semiclassical i apperantly wasnt logged in

Ted Shifrin

So write the expression for the particular solution $z_0$ (I don't like that notation) in terms of real and imaginary parts.

Mahmoud

18:07

@TedShifrin What is your explanation for $e^{\pi i}=-1$ ?

Null

sorry

Ted Shifrin

@Mahmoud: I don't know what you mean by explanation. $e^{it}$ goes around the unit circle an angle $t$.

The proof is by differential equations or Taylor series.

Mahmoud

Other than the substitution in Euler's formula. @TedShifrin

Semiclassical

Turning around and walking a step forward leaves you in the same position as taking a step backwards.

Null

@Semiclassical have I posted this more then one times?

Ted Shifrin

18:08

There's no other explanation that i know.

Semiclassical

No. Spam isn't the right word.

Ted Shifrin

Obnoxious post? :D

Null

@Semiclassical ah ok, i see what you mean!

Semiclassical

But posting one big message to chat is a bit---obnoxious, yes.

Kasmir Khaan

@TedShifrin double integral over Y dS = double integral of cross product of r' (t) * r'(s) ds dt

Null

18:09

@Semiclassical ok, i will remember that

Ted Shifrin

Magnitude of the cross product, @Kasmir. That's the "fudge factor" that accounts for how much area is distorted by the parametrization.

Kasmir Khaan

@TedShifrin but z goes from 0 to 1 ,what am having problem understanding now that this is vertical thing not horizantel like normal double integral

Steve

@TedShifrin $4\cdot \cos(t) + i\cdot \sin(t)$ ? $\Re=4$ and $\Im=1$

Kasmir Khaan

yes thanks , just didint know how to type that

Gridley Quayle

Hi, Could I get any help with 4iii?

Semiclassical

18:11

@Axoren "I should start grading this now. -Every TA ever.

Mahmoud

@TedShifrin I meant something like this .. youtube.com/watch?v=F_0yfvm0UoU (The best video on YouTube).

Gridley Quayle

Ted Shifrin

The region you integrate over is in $(s,t)$-space, @Kasmir, so it is "horizontal" — a regular old double integral.

Gridley Quayle

I think I want to show that $e_i$ divides $d_i$ and vice versa but I am not sure how the hint helps

Ted Shifrin

No, @Steve. You have to multiply everything out.

Gridley Quayle

18:12

I am also not sure how to do 4i

Kasmir Khaan

@TedShifrin OMG thanks i think i got it now ! Ted you are the best !

Ted Shifrin

$(4+6i)(\cos t+i\sin t) = $? @Steve

Yippee, @Kasmir. Keep practicing!

Kasmir Khaan

@TedShifrin will do sir ! :)

Semiclassical

Today's SMBC is pretty good: smbc-comics.com/comic/law-of-social-media

Null

can someone look over this? i feel the step goes in the wrong direction pastebin.com/tzNACpGy

Ted Shifrin

18:15

@Gridley: Don't try to leap to the end. Work it out part by part. They're leading you through the proof.

Semiclassical

So was the day before---debating whether to show my sister (the actress) that one at some point :P

Ted Shifrin

@Gridley: In general, when you see a problem like this and you freak out, try a concrete example and understand what's going on. Like for (i), take $m=4$ and $n=6$ or something.

Gridley Quayle

@TedShifrin I wasn't sure how to do 4i so I moved onto 4ii (where you had to use the previous result) which was fine.

Alessandro

Hi @Ted

Gridley Quayle

Funnily enough I did that example and you get the LHS is 0 and 3 mod 6 while the right hand side it 0 and 1 mod 2

Ted Shifrin

18:17

Hi @Alessandro :)

Wait, @Gridley. We're supposed to get $\Bbb Z/2$.

Steve

@TedShifrin Oh you meant that one. Someone told me otherwise, oh well.

$\Re((4+6i)(\cos t+i\sin t))=\Re(4 \cos t + 4i\sin t+6i\cos t-6\sin t)=4\cos t - 6 \sin t$

Is that right?

Gridley Quayle

Yeah, you get that Z/6 [4] is isomorphic to Z/2

But i couldn't find a bijection to show that was the case

Ted Shifrin

Yes @Steve.

Steve

Great, thanks :)

Ted Shifrin

Can you prove it with the fundamental homomorphism theorem, @Gridley?

Gridley Quayle

18:21

@TedShifrin is that the First Isomorphism Theorem?

Ted Shifrin

Yup.

Astyx

Hi all

Ted Shifrin

Salut, @Astyx

Astyx

Comment ça va ?

Ted Shifrin

Ça va bien, merci, et toi?

Astyx

18:23

Ça va ça va :)

Devoir surveillé d'informatique demain

Null

@Astyx salut

Astyx

Salut ! @Null

Gridley Quayle

@TedShifrin How does it apply? Z/n is a quotient group but Z/n [m] is not

Ted Shifrin

No, it's a subgroup of $\Bbb Z/n$, but you're claiming it's isomorphic to $\Bbb Z/d$.

Gridley Quayle

so do I need a surjective homomorphism from Z/n to Z/d?

Ted Shifrin

18:28

Precisely :)

Gridley Quayle

Okay thanks, what about for 4iii?

will I need to use that homomorphism again?

Ted Shifrin

No. I think they want you to see it by the count in (ii) and by symmetry considerations. I need to get going now. If you're still stuck later when I'm back, I'll ponder then.

Astyx

@Bye ted

Ted Shifrin

A bientôt :)

Daniel Fischer

18:42

@RandomVariable I don't know the decay of Bessel functions off-hand, but isn't $\dfrac{1}{\sqrt{1+x^2}} J_1(bx)$ integrable for $b > 0$?

Bon soir @Ted.

Null

Let $a\in\mathbb{R}^{+},n\in\mathbb{N}_2$.
To show: $(1+a)^n\geq\binom{n}{2}a^2$
By the binomial theorem in we have
$$(1+a)^n=1+an+\underline{a^2\binom{n}{2}}+\cdots$$
which is certainly greater or equal than $\binom{n}{2}a^2$

is this correct?

Astyx

$n\in\Bbb N_2 \iff n\ge 2$ ?

If so, yes

Null

@Astyx that's what i mean. is there a better notation for N_2?

Astyx

$n\ge 2$

Daniel Fischer

$\mathbb{N}\setminus \{0,1\}$

Astyx

18:53

To me $\Bbb N_2$ rather means $\{0,1,2\}$

Krijn

@DanielFischer Please don't start a $0 \in \mathbb{N}$ debate

Null

haha

Astyx

Is there even a debate ?

:p

Null

´n in N AND n>1

Krijn

@Astyx See math.stackexchange.com/questions/283/is-0-a-natural-number

Daniel Fischer

18:55

@Krijn The good thing is that $\mathbb{N}\setminus \{0,1\}$ means the same thing for those who (mistakenly) believe $0 \notin \mathbb{N}$ as for the others.

2

Random Variable

@DanielFischer Yes, but I'm trying to show that the function defined by the integral is an analytic function for $\text{Re}(a) >0$ by arguing that we can differentiate under the integral sign for all $\text{Re}(a) >0$. So don't we also need an integrable dominating function for the derivative of the integrand?

Null

i seriously overcomplicated my excercice

i mean, really haha

tried to induct this thing :/

Astyx

Hi @MikeMiller

Daniel Fischer

@RandomVariable If you want to argue by differentiating under the integral, yes. But if you use continuity + Morera, you don't need to.

Astyx

@Null You should always look if there is an "obvious" way of doing things ;)

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$Mathematics$ Mathematics