Mathematics - 2015-01-28 (page 4 of 4)

Pedro Tamaroff

20:12

@DonLarynx Oh?

N3buchadnezzar

@Huy DG?

Huy

diffgeo

Hippalectryon

DargonGall

N3buchadnezzar

20:28

@Huy ugh differential forms

Mike Miller

Ugh?

N3buchadnezzar

uuuuuuuuuuuuuuuuuuuuuugggggggggggggghhhhhhhhhh

@MikeMiller I have 150 problemsets to correct, any comforting words?

Mike Miller

No.

Gato

Hi all!

Studentmath

So, another question. If $G$ is finite abelian group, $o(G)=p_1^{n_1}\dots p_k^{n_k}$, $G$ would have the largest element if $G=C_{p_1^{n_1}}\times \dots \times C_{p_k^{n_k}}$?

Pedro Tamaroff

20:43

@Studentmath I don't understand your question.

"Largest element"?

Studentmath

Element of largest order, compared to other $G$'s with same order.

Pedro Tamaroff

Every finite abelian group has an element of largest order.

I still don't understand the "$G$ would have the largest element if..." part.

Ramanewbie

hi @gato !

Studentmath

@Pedro I mean that, say, $o(G)=2^23^2$. I have 4 different abelian groups of such order, correct?

Ah, I get the point.

No, wait, I don't.

Pedro Tamaroff

I think I understand what you mean.

Studentmath

20:47

Then the one with the element of largest order out of all 4 different groups, would be $G=C_{2^2}\times C_{3^2}$.

Pedro Tamaroff

Say $C_p\times C_p$ has order $2$, but the order of its elements is at most $p$.

Studentmath

The others would have an element of largest order, of smaller order than the rest.

Pedro Tamaroff

Yet $C_{p^2}$ has an element of order $p^2$.

Studentmath

Yeah.

Pedro Tamaroff

Yes, this has to do with exponents @Studentmath.

For any abelian group $G$, there is a least $n$ such that $nG=0$.

This is called the exponent of $G$. Some people call any $k$ such that $kG=0$ an exponent. So the exponent is the least integer that is an exponent of $G$.

$G$ is cyclic if and only if $n=G$.

Studentmath

20:48

Oh. That's nice

Pedro Tamaroff

Note that $|G|$ is an exponent of $G$, so $\exp G\leqslant |G|$.

@Studentmath Try to prove it. Tell me if you're stuck.

Studentmath

So a cyclic group has the largest exponent?

Pedro Tamaroff

Among groups of its order, yes.

That's what you observed.

Studentmath

Cheers, will try to do that exercise

Doesn't it come directly from the fact that the order of elements divides the order of the group?

Pedro Tamaroff

@Studentmath Just write down the proof.

You want to see that $G$ must contain an element of order $|G|$.

Show that if $g\in G$ has maximal order, it has order $\exp G$, @Studentmath.

evinda

21:02

@DanielFischer In my lecture notes, it says that we suppoe that we have a solution $x_n=\sum_{i=0}^n a_i 7^i$ of $x_n^2-2 \equiv 0 \mod{7^{n+1}}$ and then we are looking for a $x_{n+1}$ such that $x_{n+1}^2 -2 \equiv 0 \mod{7^{n+2}}$ and $x_{n+1} \equiv x_n \mod{7^{n+1}}$.
After having shown that there is a $a_{n+1}$ such that $x_{n+1} = \sum_{i=0}^{n+1} a_i 7^i$, it says that we want to have solutions for each $n \in \mathbb{N}$ (of $x^2 \equiv a \mod{p^n}$), i.e. for $n \to +\infty$ and $x_{\infty}=\lim_{n \to +\infty} x_n=\sum_{n=0}^{\infty} a_n 7^{n+1}$.

Studentmath

@Pedro wouldn't it be enough to show that $\exp G$ divides $|G|$?

Pedro Tamaroff

@Studentmath Why?

Studentmath

@Pedro because then we have that $|G|$ is an exponent, and $\exp G\le |G|$

Pedro Tamaroff

$|G|$ is always an exponent.

By Lagrange.

Daniel Fischer

@evinda I don't understand your question. You assume that your $x_n$ satisfies $x_n^2 \equiv 2 \pmod{7^{n+1}}$. The other solution of $a^2 \equiv 2 \pmod{7^{n+1}}$ is then $-x_n$, or, if you wish to write it that way, $7^{n+1} - x_n$.

evinda

21:11

@DanielFischer I haven't understood if it is right to assume that the solution of $x_n^2 \equiv 2 \pmod{7^{n+1}}$ is $x_n=\sum_{i=0}^n a_i 7^i$ or if it should be $x_n=\sum_{i=0}^n a_i 7^{i+1}$.. :/

Daniel Fischer

@evinda $\sum_{i=0}^n a_i 7^i$. It must not be divisible by $7$, so you need a $a_07^0$ term.

Studentmath

Oh, I only now got that you mean to prove that $G$ is cyclic iff $|G|=\exp G$....

Eric Gregor

Are simple functions complex-analytic?

i retract the question

evinda

@DanielFischer A ok.. So should it then be $x_{\infty}=\lim_{n \to +\infty} x_n=\sum_{n=0}^{\infty} a_n 7^n$, with $0 \leq a_n \leq 6$ ?

Daniel Fischer

@evinda Yes. I didn't notice the $7^{n+1}$ there before. That's a typo.

evinda

21:20

@DanielFischer A ok :-) Could you maybe also explain me why we don't expect that $\lim_{n \to +\infty} x_n$ is an integer?

Daniel Fischer

@evinda It can't be an integer. For every integer $k$, there is an $m$ such that $7^m > k^2$. Then you can't have $k^2 \equiv 2 \pmod{7^n}$ for any $n \geqslant m$.

Pedro Tamaroff

@Studentmath Yes. =)

evinda

@DanielFischer I am confused now.. Could you explain it further to me?

@DanielFischer Is it also related to the fact that $x^2-2=0$ has no solution in $\mathbb{Z}$ that $x_{\infty}=\sum_{n=0} a_n 7^n$ is not an integer? :/

Daniel Fischer

@evinda The only way that an integer $k$ satisfies $k^2 \equiv s \pmod{7^n}$ for all $n$ is that $k^2 = s$, since for $7^n > k^2$, the representative of the residue class of $k^2$ modulo $7^n$ is $k^2$ itself. Since $2$ is not the square of an integer, that cannot happen for $s = 2$.

@evinda It is exactly that fact.

Ted Shifrin

21:38

Hmmm, exponent of a finite abelian group was part of the first question on our algebra qualifying exam, @Pedro. And using it to prove that the multiplicative group of a finite field is cyclic. Didn't go very well.

Pedro Tamaroff

@TedShifrin Weird. It's not that complicated.

Ted Shifrin

puts on ignore anyone who says "ugh" about differential forms

Pedro Tamaroff

@TedShifrin What was the question precisely?

Ted Shifrin

Plus they know all the structure theory of finitely generated modules over a PID, @Pedro. Sigh.

robjohn

@MikeMiller: I am on campus today.

Ted Shifrin

21:40

Hi @robjohn

evinda

@DanielFischer So the solution of $x_n^2-2 \equiv 0 \mod{7^{n+1}}$ is equal to $x_n=\sum_{i=0}^n a_i 7^i$. If the latter would be an integer for all $n \in \mathbb{N}$ that would mean that $x^2-2$ has an integer solution, but that doesn't happen, so at $\infty$ we expect that $x_{\infty}$ is not an integer.
Or have I understood it wrong?

Ted Shifrin

@Pedro: Let $F$ be a finite field. (a) Tell me about the structure of $(F,+)$. (b) We defined exponent of a finite abelian group. Prove that there's an element of order the exponent. (c) Prove that $F^\times$ is cyclic.

Pedro Tamaroff

@TedShifrin Oh, OK.

Daniel Fischer

@evinda $x_n$ is an integer for all $n$. But the limit (in $\mathbb{Z}_7$) is not an integer.

Pedro Tamaroff

The proof I know regarding $F^\times$ is the one that uses $\varphi$.

Arkamis

21:43

@DanielFischer Have you gone on a meta editing spree?

Ted Shifrin

Hmm, I don't think I know that proof.

Daniel Fischer

@Arkamis Yes. Fixing borken links. I thought a little and decided it was less bad to get over with it in one go, flooded front page may go to wherever.

robjohn

@TedShifrin Hey, Ted. I am at UCLA proctoring a logic exam

Ted Shifrin

a logic exam?

evinda

@DanielFischer Do we know that $x_n$ is an integer for all $n$ since $a_n$ is an integer for all $n$ and $7^n$ too and thus also $\sum_{i=0}^n a_i 7^i$?
But if the limit is would also be an integer that would mean that $x-2$ has an integer solution?

Ted Shifrin

21:45

this is how you moonlight now, @robjohn? :P

Arkamis

@DanielFischer Ah, gotcha. I visited meta for the first time in a couple of days and was like "woooahhh what is even happening?"

Hippalectryon

@TedShifrin If we have $f(x)\sim_a g(x)$, do we always have $\displaystyle\lim_{e\to0}\int_a^{a+e}f(x)dx=\lim_{e\to0}\int_a^{a+e}g(x)dx$ ?

Pedro Tamaroff

@TedShifrin Well, one knows that over $F^\times$ the equation $x^d=1$ has at most $d$ solutions.

Ted Shifrin

Yes, I use that too, @Pedro.

Pedro Tamaroff

@TedShifrin OK.

Daniel Fischer

21:45

@Arkamis Unfortunate timing then.

Ted Shifrin

@Hippa: What's your definition of $f(x)\sim_a g(x)$?

Daniel Fischer

Meta is unusually quiet these days.

Hippalectryon

@TedShifrin $f(x)$ equivalent to $g(x)$ in $a$

Ted Shifrin

ça ne me dit rien, @Hippa

Hippalectryon

@TedShifrin Tu n'aurais pas un contre example ?

Pedro Tamaroff

21:47

@TedShifrin Say $F$ has order $q$, then $G=F^\times$ has order $q-1$, and every element has $x^{q-1}=1$. This splits completely into linear factors, and for any divisor $d$ of $q-1$ you know $x^d-1$ divides $x^{q-1}-1$ so it has distinct roots.

Ted Shifrin

Je n'arrive pas à comprendre ce que tu veux dire @Hippa

Hippalectryon

@TedShifrin Quele partie ?

Ted Shifrin

La définition

Mike Miller

Sounds awful, @DanielF. How will you cope?

robjohn

@TedShifrin I logged in during a quiet time. Proctoring is not the most exciting of pasttimes

Ted Shifrin

21:48

So is this what you do on your time off from your other job, @robjohn? Or are you still at UCLA in some capacity?

Ramanewbie

anyone knows autoit ?

Hippalectryon

@TedShifrin De l'équivalent ? $f(x)\sim_a g(x)$ lorsque il existe $\epsilon(x)$ telle que $\epsilon(x)\to_a 0$ et $f(x)=g(x)+\epsilon(x)g(x)$

Daniel Fischer

@MikeMiller Handle a couple of flags, idle around in chat, the usual stuff.

Ted Shifrin

eh bien, $f-g = o(g)$.

$g$ continuous on $[a,a+\delta]$?

Hippalectryon

Uh pas sur. En tout cas, oui sur $(a,a+\delta)$

Don Larynx

21:50

Hi @Ted, give me three numbers, each below 2.147 *10^9.

They can be negative.

Ted Shifrin

Mais les intégraux convergent?

Studentmath

Think I wrote it down right, Pedro. And I just found an intriguing question, gonna try it out.

Hippalectryon

@TedShifrin intégraux ?

Ted Shifrin

oops, intégrals

hmm

Mike Miller

@Daniel Why'd ya bump 200 meta questions?

Ted Shifrin

21:51

non, j'avais raison :P

robjohn

@TedShifrin I am the developer for the software that is used at UCLA (and about 2 dozen other universities) to teach first year sentential logic. Part of what I do is to proctor the exams here.

Hippalectryon

@TedShifrin oui, elles convergent. Et non, intégraux n'existe pas.

Ted Shifrin

oops, intégrale ... j'oublie trop

Daniel Fischer

@MikeMiller Because. And the way you count, don't play Poker.

Ted Shifrin

ah, interesting @robjohn ... distance learning stuff, then? There are mass sentential logic courses?

applauds @DanielF :D

Alors, @Hippa: $\int_a^{a+\delta} \epsilon(x)g(x)\,dx \le \epsilon \int_a^{a+\delta}g(x)\,dx$ if $\delta$ is small enough. So, yes?

robjohn

21:54

@TedShifrin No, the course is being taught now from a text that is written around the software. There are courses at a number of universities.

Hippalectryon

@TedShifrin Thanks

Ted Shifrin

I guess I'm shocked that so many universities teach such a course. To what audience, @robjohn?

Pedro Tamaroff

@TedShifrin Sorry Ted I didn't finish my proof. This means the subgroup $\{x:x^d=1\}$ has $d$ de elements, so if you take $\psi(c)$ to be the number of elements of order $c$ in $G$, then $\sum_{c\mid d}\psi(c)=d$. Inverting gives $\psi(c)=\varphi(c)$.

Taking $c=q-1$ gives an element of order $q-1$.

Mike Miller

I just got the ping, @robjohn. What are you up to now?

@DanielF Well, I play with other math students, and they seem to be worse at counting.

Ted Shifrin

but the old gay guys can count, @Mike? :D

Mike Miller

22:02

None of them are in combinatorics, luckily.

Ted Shifrin

I tell people at cocktail parties that mathematicians are no better at balancing checkbooks than other mortals :)

Arkamis

@TedShifrin I do the same. "Oh, you're a mathematician. You must be good at numbers." "Ummmmm"

(the sad part is I'm a numerical analyst)

Ted Shifrin

so you need to be better at numbers than I do :P

robjohn

@TedShifrin The software had been written around this book, but my program made it easier to teach and so a new book was written.

@TedShifrin It is mostly for undergrads, but there are some grads who have not had the proper background in logic for their philosophy classes

Ted Shifrin

@robjohn: Is this an intro course for CS majors? A math for poets type of course? I just don't see the audience.

robjohn

22:06

@MikeMiller finishing up the exam now, there are some stragglers who require a bit more time.

Ted Shifrin

Hmm, weird. I've never seen a course like this any place I've been. We have a discrete math for CS course, but it has a lot more content than just sentential logic.

Mike Miller

@robjohn I don't have anything planned until 4, if you'd like to grab coffee.

robjohn

@TedShifrin No, it is part of the philosophy dept

Ted Shifrin

oh, we have a symbolic logic course in the philosophy department. I guess that's what this is, sorta. I get it. Thanks.

robjohn

@MikeMiller We will probably be here another half hour. I will let you know when we have finished.

Ted Shifrin

22:07

damn, @Mike gets to meet @robjohn before I do :(

Mike Miller

Sure, @robjohn.

Ted Shifrin

@Pedro, once we have the exponent, we're done by your early comment. Because if the exponent $<d=q-1$, then we have $d$ roots of a polynomial less than $d$. Done.

Pedro Tamaroff

@TedShifrin That's a very nice argument.

Good.

Kevin Driscoll

@TedShifrin Logic courses get weird as you get to the higher level. I took a class on Modal Logic that was sorta sprinkle in the set theory and the topology and the analysis and all sorts of bits and bobs.

Ted Shifrin

I took a graduate course in logic as an undergraduate. I loved the model theory stuff (including hyperreal numbers), but after weeks of Turing machines, gave up and dropped the course.

Don Larynx

22:19

@Pedro I proved that, in the division algorithm, the term $r_n$ contains $f(n)$ number of $a$'s and $f(n+1)$ number of $b$'s. For example, when $r_2 = 0$, and $gcd(a, b) = r_1$, then we have $a - bq_1 = r_1$ so it has 1 $a$ term and 1 $b$ term. Similarly, when $gcd(a, b) = r_2$, $$b - (a-bq_1)q_2 = r_2$$ so we have 2 $b$ terms and 1 $a$ terms. We continue inductively to prove our claim, $r_n$ contains $f(n)$ number of $a$'s and $f(n+1)$ number of $b$'s

Pedro Tamaroff

What is $f$?

Don Larynx

The Fibonacci sequence

i.e. when $r_{13} = gcd(a, b)$ we have $f(13)$ a's and $f(14)$ b's

Ramanewbie

I am given $A\left(−8;3\right)$, $B\left(−6;−6\right)$, $C\left(−7;2\right)$.

I want to calculate the equation of the median from C.

If M is the point such that $\left(MC\right)$ is the median from C,

$MC : y=mx+p$

$m=\dfrac{\frac{y_a+y_b}{2}-y_c}{\frac{x_a+x_b}{2}-x_c}$, So

$m=\dfrac{\frac{y_a+y_b}{2}-y_c}{\frac{-8-6}{2}+7}$, So

$m=\dfrac{\frac{y_a+y_b}{2}-y_c}{0}$

And that's impossible ! Although, I now this equation exists... Where does it go wrong is what's just above ?

Ted Shifrin

What is the midpoint of $\overline{AB}$, @Ramanewb?

Ramanewbie

@ted "midpoint" ?

@ted oh ok

Ted Shifrin

22:24

la moitié de $A$ à $B$ ?

Ramanewbie

@ted hum what's the point of the overline on $AB$ ?

Ted Shifrin

for me, $AB$ means the length, $\overline{AB}$ means the line segment

Ramanewbie

@ted ok, that must be the american way...

Ted Shifrin

it's not universal here, either

Ramanewbie

ok

@ted it's

$x(midpoint_AB)=\dfrac{x_a+x_b}{2}$ and $y(midpoint_AB)=\dfrac{y_a+y_b}{2}$.

Ted Shifrin

22:27

OK, et puis?

Oh, I see the problem. The $x$-coordinates are the same. So the line is not of the form $y=mx+b$ !!

Ramanewbie

@ted they are the same ?? for what points ?

Hippalectryon

if anyone knows...

Ramanewbie

@hippa knows what ?

Ted Shifrin

for $C$ and the midpoint, @Ramanewb

@Hippa: Antonio Vargas just added a comment with a reference for you.

Ramanewbie

@ted yes, I get that.

@ted the $x$-coordonates are the same so it's not $y=mp+p$, right ?

Ted Shifrin

22:31

right

Ramanewbie

@ted so what ?

Ted Shifrin

How do you give the equation of a vertical line?

Ramanewbie

@ted I'm stupid, of course !

Ted Shifrin

c'est ton frère qui dit ça souvent :P

NON @Ramanewb

Ramanewbie

@ted lol...

@ted no WAIT !!!!!

$x=k$ I mean ! @ted

Ted Shifrin

22:34

beaucoup mieux

Ramanewbie

@ted I'm just tired ^^ or not...

so $k=-7$ right @ted ?

Ted Shifrin

bien sûr

evinda

@DanielFischer According to my lecture notes, the representation of the integer p-adic $x=(x_n)_{n \in \mathbb{N}_0} \in \mathbb{Z}_p$ is called "opened" when $0 \leq x_n<p^{n+1}$. But doesn't this always hold?

Ted Shifrin

non @Ramanewb

Ramanewbie

??

Daniel Fischer

22:36

@evinda That depends on the representation. There are many many many ways to define $\mathbb{Z}_p$.

Ramanewbie

$x+7$ I mean@ted

Ted Shifrin

Trop de fautes, @Ramanewb. Bonne nuit!

Ramanewbie

@ted lol...

robjohn

@MikeMiller Kerckhoff patio?

Ramanewbie

@ted it's too late, I can't think properly !

Mike Miller

22:38

Sure, @robjohn. Give me a minute to drop stuff off in my office.

Ramanewbie

that's the only reason I do so many mistakes... @ted

evinda

The definition is this: $$\mathbb{Z}_p=\{ (\overline{x_n})_{n \in \mathbb{N}_0} \in \prod_{n=0}^{\infty} \mathbb{Z}/p^{n+1}\mathbb{Z} | x_{n+1} \equiv x_n \mod{p^{n+1}}\}$$ @DanielFischer

Ted Shifrin

Have a drink for me @robjohn @Mike

Hippalectryon

@Ramanewbie go to bed then

Ramanewbie

@hippa Sure ! That's what I'm doing, actually !

Daniel Fischer

22:40

@evinda Then you don't necessarily have $x_n < p^{n+1}$, since $x_n$ is not determined by $\overline{x_n}$.

robjohn

@MikeMiller I am holding a Logic Team hat :-)

evinda

@DanielFischer $$(\overline{x_n}) \in \prod_{n=0}^{\infty} \mathbb{Z}/p^{n+1}\mathbb{Z}$$

means that $(x_1, x_2, \dots, x_k, \dots )$ where $x_k \in \mathbb{Z}/p^{k+1} \mathbb{Z}$, right?

Doesn't $x_k \in \mathbb{Z}/p^{k+1} \mathbb{Z}$ mean that $x_k$ is a residue class modulo $p^{k+1}$, i.e. that it is an integer from the interval $[0, p^{k+1}-1]$ ? Or have I understood it wrong?

Daniel Fischer

@evinda No, $x_k \in \mathbb{Z}$, but $\overline{x_k} \in \mathbb{Z}/p^{k+1}\mathbb{Z}$.

evinda

@DanielFischer So you mean that $(\overline{x_k})=(\overline{x_1}, \overline{x_2}, \dots)$, right? But then how is $x_k$ related with the definition?

Daniel Fischer

@evinda That's the snag in that definition. The $x_k$ are not (uniquely) determined by $x\in \mathbb{Z}_p$. Only the $\overline{x_k}$ are.

evinda

22:56

@DanielFischer I haven't understood which will then be the form of $x_k$... :/

Daniel Fischer

$x_k$ is just an ordinary integer. The sequence of the $x_k$ needs to satisfy the compatibility relation $x_{k+1} \equiv x_k \pmod{p^{k+1}}$, but that's all you generally know about the $x_k$.

evinda

@DanielFischer So does it have to hold that $(\overline{x_k}) \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ and $x_{k+1} \equiv x_k \mod{p^{k+1}}$ ? I thought that the same component has to satisfy both of the above relations... Am I wrong?

I am confused now.. Do we maybe relate the two definitios of $\mathbb{Z}_p$ and $x_n$ is equal to $x_n=\sum_{i=0}^n a_i p^i$ ? :/

Daniel Fischer

@evinda It is convenient to take $x_n = \sum_{i=0}^n a_i p^i$. But it is not necessary to take those. You can add arbitrary multiples of $p^{n+1}$ to $x_n$ if you wish. (You shouldn't wish, but meh.)

evinda

23:17

@DanielFischer So we have a $x \in \mathbb{Z}_p$ with $x=\overline{x}_n=(\overline{x_0}, \overline{x_1}, \overline{x_2}, \dots, \overline{x_k}, \dots)$ and $\overline{x_k} \in \mathbb{Z}/p^{k+1}\mathbb{Z}$. And we say that $x$ is "opened" if $0 \leq x_n< p^{n+1}$. Could you explain me how we relate the two definitions of $\mathbb{Z}_p$? I am confused now..

Studentmath

I have another question.

Hippalectryon

@TedShifrin I'm afraid the comment on math.stackexchange.com/questions/1124162/… doesn't really help.

@TedShifrin His method involves an integral which I cannot compute.

@MikeMiller Got an idea ?

The point that's important is the equivalent near $1$. I don't know where D'aurizio got it.

@Chris'ssis Maybe you know :D

Chris's sis

I couldn't sleep, I have to ask that!

Why on earth Mathematica gives different results here?

NIntegrate[E^(-(x + y)), {x, 0, Infinity}, {y, 0, Infinity}]=1

and

NIntegrate[E^(-(x + y)), {x, 0, 0.1}, {y, 0, 0.1}] + NIntegrate[E^(-(x + y)), {x, 0, 0.1}, {y, 0, Infinity}] + NIntegrate[E^(-(x + y)), {x, 0, Infinity}, {y, 0, 0.1}] + NIntegrate[E^(-(x + y)), {x, 0.1, Infinity}, {y, 0.1, Infinity}]=1.01811

Hippalectryon

Because NIntegrate = approximation, I guess

Chris's sis

Indeed, I'm exceptionally tired, but still, I don't see why.

@Hippalectryon I took that N out

Integrate[E^(-(x + y)), {x, 0, 0.1}, {y, 0, 0.1}] + Integrate[E^(-(x + y)), {x, 0, 0.1}, {y, 0, Infinity}] + Integrate[E^(-(x + y)), {x, 0, Infinity}, {y, 0, 0.1}] + Integrate[E^(-(x + y)), {x, 0.1, Infinity}, {y, 0.1, Infinity}]=1.01811

Maybe I did something incorrectly?

Hippalectryon

23:32

Are the limit swaps all legal ?

@Chris'ssis So, $\displaystyle\int_0^{.1}\int_0^{.1}+\int_0^\infty\int_0^{.1}+\int_0^{.1}\int_0^‌\infty+\int_{.1}^\infty\int_{.1}^\infty$ ?

Chris's sis

@Hippalectryon I check that again ...

Hippalectryon

@Chris'ssis Some bounds seem to overlap

Chris's sis

@Hippalectryon Yeah, typos. Damn ... I run to sleep. Thanks!

Middle integrals were the problem due to the typo.

Hippalectryon

Good night :-)

Transcript for

$Mathematics$ Mathematics