Mathematics - 2014-06-12 (page 1 of 3)

masfenix

00:00

@KevinDriscoll Thanks. What is the intuition behind Laplace Transforms? What do they do to a function? I don't understand why the condition of decay (ie, function must be of exponential order) is needed.

Kevin Driscoll

@masfenix Well if it grows faster than exponentially then the transform doesn't exist for any $s$

Pedro Tamaroff

@KevinDriscoll YO YO

Kevin Driscoll

The intuition though is that the Laplace domain is one in which differentiation by $t$ is mapped ot multiplication by $s$ @masfenix

What it does to functions though..... I can't say for sure. It's more clear with Fourier and Mellin. I don't work much with Laplace @masfenix

masfenix

Oh, I think I get it now. The decay condition is needed for the integral to converge. If the integral dosn't converge then obviously we don't have a Laplace transform.

Kevin Driscoll

Indeed.

@PedroTamaroff Yo! Good to see you.

Pedro Tamaroff

00:06

@KevinDriscoll What's up?

Kevin Driscoll

@Pedro Working on my PhD thesis proposal. And crying on the inside at how few jobs are available in my field.

Pedro Tamaroff

@KevinDriscoll Ah, cannot be less than those for a mathematician, for sure!

Kevin Driscoll

@Pedro You'd be surprised. The problem is physics is more segmented. Whether you do theory or experiment you're automatically disqualified from the other's job postings.

I think there are also fewer math PhDs. But I know its also pretty bad over there. From my cursory estimates and personal experience I'd say its about even.

Pedro Tamaroff

@KevinDriscoll Hehe, I'll tell you when I graduate.

Kevin Driscoll

@pedro I'm hoping to not wait until I'm ready to graduate to really be looking for things

Pedro Tamaroff

00:22

@KevinDriscoll But you're already a PhD student right?

@AlexanderGruber HAI.

Alexander Gruber

@PedroTamaroff Bon soir.

Kevin Driscoll

@Pedro Yeah, so a good bit farther along. Seems like a lot of people don't start looking for jobs until they graduate tho.

Pedro Tamaroff

@AlexanderGruber What have you been up to?

Alexander Gruber

@Kevin oh you're a physics PhD student?

@PedroTamaroff torturing my business calc students :)

Kevin Driscoll

@AlexanderGruber Yup. Cold atom theory.

Alexander Gruber

00:24

@KevinDriscoll sounds fun. i like physics, did my undergrad in it

i'm not quite awake yet.

Kevin Driscoll

@AlexanderGruber Did you also do math?

Alexander Gruber

@KevinDriscoll yeah, doubled up

Kevin Driscoll

AH okay. I wish I had done math sometimes.

Alexander Gruber

@KevinDriscoll did you during undergrad?

@PedroTamaroff yourself?

Kevin Driscoll

Nope. I was a physics/philosophy double @alexander

Alexander Gruber

00:27

@KevinDriscoll Ahhh interesting.

So you can ponder the ethics of relativity

Kevin Driscoll

Hahahaha. Well it all depends on you frame of reference really.......

2

Pedro Tamaroff

@AlexanderGruber Algebrain' pretty hard.

I failed my combinatorics test as I expected. =P

Kevin Driscoll

@pedro you failed a test on combinatorics? Howd that happen?

Pedro Tamaroff

@KevinDriscoll A series of bad choices and mistakes during the exam.

Kevin Driscoll

@pedro What topics did it cover?

Pedro Tamaroff

00:38

Elementary Counting, Generating Functions, Graph Theory, Incidence Algebras of posets.

Kevin Driscoll

@pedro I probably would fail that test too. Generating functions are aparticular weak point for me. Though we use a generalized version in Quantum Field Theory

G.T.R

05:25

@Hippalectryon $\mathbb Q$ Lacks Bolzano-Weierstrass property

Balarka Sen

06:33

@mick I can't talk about magic floopskintunks and say that they make sense but I can't explain, right?

First show me how you define exp(X) in R[X] modulo X^3 - 1. You are not even sure whether or not it makes sense, i.e., is a polynomial.

Hey @Alexander.

Alexander Gruber

hi

real empty in here tonight

Balarka Sen

empty?

David Kirby

∅

Balarka Sen

Heh.

I couldn't find any stuff in Isaacs, by the way @Alexander

David Kirby

@Balarka Sen, I've been on a quintic trip for the past few days thanks to you. Teehee

Balarka Sen

06:42

@DavidKirby Oh? Tell me about it.

@Alexander I totally not approve the weird man mustache.

Alexander Gruber

@BalarkaSen hahaha

i grew it out specifically for my ID picture

i don't normally have facial hair

David Kirby

@Balarka Sen, well, I've been working a lot with Harmonic Series lately and have come up with notebooks full of interesting series. After talking about quintic equations with you I started trying to solve various series with a quintic equation on the right hand side, as well as trying to rewrite different types of series in quintic form and the results of have been really interesting.. Roots galore, which I guess is to be expected...

Balarka Sen

@Alexander To be a mathematician, you need to look like this

bad pun

@DavidKirby Examples?

David Kirby

@Balarka Sen, sure, gimme a sec and I'll type in an example

Ok, here's one -- $$H_{-x-\frac{3}{4}}-H_{-x-\frac{1}{4}}=x^5-x-Z$$

Solving for Z gives a really interesting result :

$$Z\to \frac{1024 H_{-x-\frac{1}{4}}-1024 H_{-x-\frac{3}{4}}-1024 \left(-x-\frac{1}{4}\right)^5-1280 \left(-x-\frac{1}{4}\right)^4-640 \left(-x-\frac{1}{4}\right)^3-160 \left(-x-\frac{1}{4}\right)^2+1004 \left(-x-\frac{1}{4}\right)+255}{1024}$$

Balarka Sen

Whoa.

That's odd.

David Kirby

06:56

yeah now check this :

$$H_{-x-\frac{3}{4}}-H_{-x-\frac{1}{4}}=\frac{-1024 H_{-x-\frac{1}{4}}+1024 H_{-x-\frac{3}{4}}+1024 \left(-x-\frac{1}{4}\right)^5+1280 \left(-x-\frac{1}{4}\right)^4+640 \left(-x-\frac{1}{4}\right)^3+160 \left(-x-\frac{1}{4}\right)^2-1004 \left(-x-\frac{1}{4}\right)-255}{1024}$$

solving for x :

$$\{\{x\to -1\},\{x\to 0\},\{x\to -i\},\{x\to i\},\{x\to 1\}\}$$

Balarka Sen

Ah, so you have a solvable quintic up there.

Let me understand what you are trying to say,

$x^5 - x - \tau = \mathcal{H}_{-x-3/4} - \mathcal{H}_{-x-1/4}$

When $\tau$ is the humongous thing above, right?

David Kirby

Correct, ... my thinking was that this is the expected behavior though, right?

Balarka Sen

So shouldn't the harmonic number terms cancel out, leaving you some algebraic identity?

David Kirby

Yup, it reduces to : $$-H_{-x-\frac{1}{4}}+H_{-x-\frac{3}{4}}-x^5+x$$

Balarka Sen

How do you find it interesting?

David Kirby

07:06

Well, it just hit me by surprise, and it's a relationship that I only understand in a one-sided sense.. I think I need to study up on Galois groups (and cyclic permutations especially) .. Not sure where to begin

Balarka Sen

How do you even get that identity? Seems weird to me.

I never imagined a connection of quintics with harmonic numbers.

Could harmonic numbers then be inverted to solve a quintic?

David Kirby

Heh, neither did I.. It was the weirdest thing, tried it on a whim after talking to you the other night.

Balarka Sen

Like, asking to solve $\mathcal{H}_x = \tau$ for $x$.

David Kirby

@Balarka Sen, I was wondering that, but hadn't figured out how to go about it yet

Balarka Sen

How is $\tau$ related to $x$ in the above quintic identity?

$$x^5 - x - \tau = \mathcal{H}_{-x-3/4} - \mathcal{H}_{-x-1/4}$$

@David

David Kirby

07:16

like this? $$\tau =H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}+x^5-x$$

uhm, err

and solving for $\tau$ gives you that big equation

and the big equation reduces back down to $$\tau =H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}+x^5-x$$

it's got my brain all blended for sure heh

Balarka Sen

Ah. But that's not weird at all, is it?

We just have another algebraic form for $x^5 - x$

And the solutions you get for $x$ are really the solutions of $x^5 - x = 0$

David Kirby

Yup, -- so that definitely makes sense.. Ok, so I'm still a little baffled by the solution after substitution :
$$H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}=H_{-x-\frac{1}{4}}-H_{-x-\frac{3}{4}}+x^5-x$$

$$\{\{x\to -1\},\{x\to 0\},\{x\to -i\},\{x\to i\},\{x\to 1\}\}$$

so instead of the 5 roots of the quintic, you get that

Balarka Sen

Yes, the roots of $x(x^4 - 1) = 0$, the simplest kind of quintic you can think of.

David Kirby

AH

Hah, ok it all becomes clear to me now

... I have fooled myself with slight-of-hand

Balarka Sen

07:34

I am trying to prove that a quintic cannot be solved in terms of polygamma and polylogarithms, based on Murty-Saradha-Gun conjecture about their algebraic independence.

Essentially tracing steps of T. Chow, who proved it for exponential and logarithms.

(assuming Schanuel's conjecture)

David Kirby

If so, it could be a boon to transcendental number theory

Balarka Sen

Probably. I've never heard of any result regarding the fact that polylogarithmically-closed of polygamma-closed fields only contain algebraic with solvable galois group.

Studentmath

07:53

David, I have a probability question I think you will enjoy

I am enjoying it, at least

Studentmath

08:10

Hrm, unsure..

David Kirby

@Studentmath, .... cuuuurrrious

Studentmath

You have a box with 6 coins, numbered 1 to 6. Each has a probability of 1/i (where i is its number) to land Head up. You have a fair dice, with 6 sides.
You throw the dice once, pick up the coin according to the number that it landed on, and throw it until you land Head up.

Let X be the RV representing the result of the dice, and Y be the RV representing the number of times you throw the coin until (including) it lands Head up.

E[Y], Var(Y), and $P(Y\ge 5$| X is even)

It's really nice, you get to toy with few known probabilities but in a nice controlled and adjusted enviroment

Chris's sis

08:29

Greetings

@robjohn Is there a known form of $\zeta''(2)$ expressed in terms of known constants?

@robjohn I wonder what is the best place to publish such a discovery (just in case).

robjohn

@Chris'ssis I think so... let me look

robjohn

08:47

@Chris'ssis Heh... I entered Sum[(Log[k]/k)^2, {k, 1, Infinity}] into Mathematica and got a useless result :-)

David Kirby

@Studentmath, interesting.. i gotta jet right now, but i'll be back later on this morning and will tinker with that a bit more

Chris's sis

@robjohn I have to go now. Pls let me know if you find something about it.

Balarka Sen

09:35

@Chris'ssis Unlikely.

Chris's sis

@BalarkaSen A friend of mine said he discovered such a thing ... (I need to see to believe)

Balarka Sen

$\zeta'(2)$ has a closed form though. Interesting question.

@Chris'ssis Try arxiv for a preprint, as an initial publication.

Chris's sis

@BalarkaSen If someone like me discovered such things, most probably would be accussed by intellectual property theft or something like that. I can guess many would claim it's their discovery (since I have no math background).

Balarka Sen

@Chris'ssis Your works are original, no? So why would anyone accuse you of thieving?

Chris's sis

@BalarkaSen Hypothetically speaking ...

Balarka Sen

09:41

You're being silly.

Nowadays, it doesn't matter if you don't have a math background.

People publish there work and none will claim it's theirs.

Chris's sis

@BalarkaSen Maybe ...

Balarka Sen

If your works are original and of general interest, publish them!

It's far better to publish research works than publishing a contest book.

@robjohn

Given any formal power series $f_1,...,f_n$ in $t\Bbb C[[t]]$ which are linearly independent over $\Bbb Q$, then the field extension $\Bbb C(t,f_1,...,f_n,\exp(f_1),...,\exp(f_n))$ has transcendence degree at least $n$ over $\Bbb C(t)$. [Ax-Schanuel theorem]

What the heck is $t\Bbb C[[t]]$?

robjohn

@BalarkaSen I assume complex power series without constant term

Balarka Sen

@robjohn $\Bbb C[[t]]$ is the ring of formal power series, but I don't understand what's the $t$ sitting up in the front.

robjohn

@BalarkaSen it means multiply by $t$ to get rid of the constant term

Balarka Sen

09:51

Weird notation for that.

@robjohn OK, thanks. That makes sense.

But wait a minute. The aren't $f_i$s linearly dependent over $\Bbb C(t)$? All having a factor of $t$?

N3buchadnezzar

Heyall

Balarka Sen

Ahoy.

N3buchadnezzar

How can I use that $\sqrt{1-t} \leq 1-t/2$ holds for $t \in (0,1)$ to prove that
$$\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$$?

Balarka Sen

Yuck.

Back to Ax-Schanuel : Doesn't the linear independence of $f_i$s over $\Bbb C(z)$ mean we are excluding the $n$-tuple candidate $\{f_i\}$ outright?

Plus, excluding all $n$-tuples with multiple $f_i$s in it?

Daniel Fischer

1

Q: Using $\sqrt{1-t}\leq 1-\frac t2$ to show that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$

I have a problem that tells me to use that $\sqrt{1-t}\leq 1-\frac t2$ for $t\in(0,1)$ to show by induction that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$ So far I have shown that $\frac{1\cdot3\cdot5\cdots(2n-1)}{1\cdot2\cdot3\cdots n}\leq 2^n$ from a previou...

induction

N3buchadnezzar

10:06

Cheers

Balarka Sen

Should I jettison the dead weight then (i.e., @N3buchadnezzar)?

N3buchadnezzar

?

Balarka Sen

@N3buchadnezzar We are in a pirate ship. Didn't I say "Ahoy"?

Chris's sis

@N3buchadnezzar I think you wanna see my proof ...

N3buchadnezzar

@BalarkaSen yarr

Balarka Sen

10:09

That sum looks familiar.

here

Note that the identity below $(3)$ is weirdly resembles the Chudnovsky brother's formula, @Chris'ssis @N3buchadnezzar

These are called Ramanujan-Sato series. Who knew?

So rigorous, much results, very identities, wow.

Studentmath

There has to be a simpler way to do this.. sigh

Any probability lads around? Nothing too complicated, just trying to figure out a shorter way to do something

Balarka Sen

10:29

@Chris'ssis Wiki says Ramanujan knew something to compute these series we don't. Interesting, isn't it?

Chris's sis

@BalarkaSen Indeed. (I hope to be like Ramanujan one day)

robjohn

@Chris'ssis There are claims that he had alien help :-)

4

Balarka Sen

I reckon Ramanujan must've thought about modular forms and the surrounding algebraic theories long before we did.

Chris's sis

@robjohn lol :-)

Balarka Sen

As I always say : Ramanujan is the father of modular forms. He knew much than we think he did.

Perhaps he created them exactly with the same rigor we introduce these days.

N3buchadnezzar

10:31

@DanielFischer That's funny, the exact same guy asked the same question on a different site. (Norwegian). Seems the problem with your explenation (which I found exellent) is that he seems to miss the crucial fact that $\prod_{k=1}^n k =\prod_{k=2}^n k = n!$

Daniel Fischer

Hmm. How?

Balarka Sen

@Chris'ssis Well, it's never late to be like a classical hero in mathematics.

Chris's sis

True

N3buchadnezzar

@DanielFischer Well it seems he does not understand why the product equals $1/\sqrt{n}$

Balarka Sen

I sure do hope to be like Erdos someday. But alas, I don't think I can ever be like him.

N3buchadnezzar

10:34

I hope to be like Galois one day

Daniel Fischer

@N3buchadnezzar Yes. I wonder how that is possible.

Balarka Sen

@N3buchadnezzar Whoa.

He was another guy like Ramanujan.

N3buchadnezzar

Swimming in p***y, and death by manly duel + mathematician.

Daniel Fischer

@N3buchadnezzar Pah, better be like Ian Curtis. Or the Lizard King.

Balarka Sen

@N3buchadnezzar Ah.

@DanielFischer Lizard? Curtis Connors?

Daniel Fischer

10:37

@BalarkaSen James Douglas Morrison

Balarka Sen

Don't know who that is.

K, I gotta go.

skullpatrol

later pal

Daniel Fischer

@BalarkaSen When you return: en.wikipedia.org/wiki/James_Douglas_Morrison

N3buchadnezzar

@DanielFischer There was an idea of making a nirvana guitar hero game, where the controller would be the shape of a shot gun..

skullpatrol

..not exactly sympathetic imo

Daniel Fischer

10:45

@N3buchadnezzar That could be considered a little in bad taste.

N3buchadnezzar

Yeah, although I would not say no to more nirvana

skullpatrol

sometimes suicide is the only way to relieve the pain.

Chris's sis

@robjohn the solution to the integral involving the dilogarithm

robjohn

@Chris'ssis why are you writing $i^2$ instead of $-1$?

Chris's sis

@robjohn it looks nice that way ( the negative result annoys me)

skullpatrol

10:52

@Chris'ssis Why? Haven't you accepted negative numbers yet :-)

Chris's sis

@skullpatrol :-))))))))

robjohn

@skullpatrol you've heard of the double negative, this is the double imaginary.

skullpatrol

:D

N3buchadnezzar

I do not believe in negative numbers

Studentmath

@N3b what about non-integers?

robjohn

10:54

@DanielFischer I feel so old.

Daniel Fischer

@robjohn Why that?

robjohn

@DanielFischer people here don't know about The Doors...

N3buchadnezzar

._.

skullpatrol

@N3buchadnezzar Do you believe in zero as a number?

N3buchadnezzar

nah

Daniel Fischer

10:57

@robjohn That's not age, that's lack of education. Like if they didn't know about Marlowe (both).

N3buchadnezzar

< 3 Riders on the Storm

robjohn

@DanielFischer It has to do with when people grew up and what music was playing at the time.

Studentmath

I grew up with hip hop nods

Daniel Fischer

@robjohn Good music is timeless.

Studentmath

There has to be an easier way to do Var(Y), I can't believe I actually have to do all these $Cov(Y_i, Y_j)$. I'd like to think they are all zero, but I can't find a good explaination. So I guess they aren't. Which means, I am stuck.

N3buchadnezzar

11:00

@DanielFischer Cream?

Studentmath

They are exclusive, so they can't be independent..

Daniel Fischer

@N3buchadnezzar Not so much my thing. Good, but not with the necessary je ne sais quoi.

Chris's sis

►

N3buchadnezzar

Creedence?

skullpatrol

When people say "grew up with;" that means, to me, the popular music when they were 13 to 19 years old.

Daniel Fischer

11:02

@N3buchadnezzar Definitely.

N3buchadnezzar

=D

robjohn

@skullpatrol That's about right, but for me, I think it was until 26 (I listened to and still really enjoy music from when I was in grad school and a couple of years after)

N3buchadnezzar

I love me some good bach

skullpatrol

Indeed, music is a very personal taste...

...like the music of the spheres :D

Hippalectryon

Hello there !

skullpatrol

11:07

Hi!

Chris's sis

@robjohn in my proof there is a text that I forgot to remove ... "By letting $x\mapsto -x$ in $X$, we get"

Hippalectryon

http://math.stackexchange.com/questions/397097/uniform-convergence-of-sum-n-1-infty-frac-sinn-x-sinn2-xnx2/398085#398085 why do we have $\begin{align}
\sum_{n = 1}^N{\sin{(nx)}\sin{(n^2 x)}\over n + x^2} & = \sum_{n = 1}^N {1\over n+x^2}(S_n - S_{n-1}) \\
\end{align}$ ?

($S_n = S_n(x) = \sum\limits_{k = 0}^n \sin{(kx)}\sin{(k^2 x)}$)

robjohn

It may have something to do with $\sin(a)\sin(b)=\frac12(\cos(a-b)-\cos(a+b))$

Chris's sis

@Hippalectryon Isn't it related to the summation by parts?

Hippalectryon

@Chris'ssis it is

Chris's sis

11:14

@Hippalectryon OK :-)

Hippalectryon

@Chris'ssis but it comes after that part

Balarka Sen

Music is noise.

robjohn

@BalarkaSen how long ago did you sell your soul? ;-)

8

Balarka Sen

a couple of years ago, i believe.

Hippalectryon

Don't worry you can get a cheap new one on ebay :3

Balarka Sen

11:17

wait, what?

@Hippalectryon this is star-worthy

Hippalectryon

:)

'Mathematician looking for a soul - cheap - good quality - science resistant - contact me anytime'

robjohn

@Hippalectryon no denomination specified?

Balarka Sen

Who starred?

You fool.

Parth Kohli

I did.

Balarka Sen

@ParthKohli I know you did.

Parth Kohli

11:20

@BalarkaSen only robjohn's.

skullpatrol

@robjohn But a zero denominator is not allowed ;-)

Balarka Sen

Wait, 3 stars?

Hippalectryon

@robjohn I'm not used to doing this - i already stole my 30 classmate's souls :D

Balarka Sen

@Hippalectryon And now you're selling them cheap?

robjohn

@BalarkaSen gotta keep the product moving

Balarka Sen

11:22

Just don't make a horcrux outta those, that's all.

Hippalectryon

@BalarkaSen Exactly, i need money for an ice cream :D

Once i have my ice cream, i multiply it into 2 ice creams thanks to the axiom of choice

And therefore i have infinitely many ice creams

Balarka Sen

No, it's Banach Tarski.

Hippalectryon

Isn't it based on the axiom of choice ?

Balarka Sen

yeah, but it's nonobvious.

Hippalectryon

Hey i'm not telling how i multiply ice creams to everyone :P

Balarka Sen

11:23

Haha

skullpatrol

If I take your ice cream away; is that negative ice cream?

Hippalectryon

@skullpatrol :O Don't ! nuuu my ice cream - hey wait, i'll just need to buy something to take the absolute value of ice creams :D (Unless it creates negative entropy duh)

Balarka Sen

You should bye a comathematcian, @Hippa, if you're interested in turning cotheorems into ffee.

Hippalectryon

@BalarkaSen HTCPCP is enough for me :)

Balarka Sen

?

Hippalectryon

11:27

Hyper Text Coffee Pot Control Protocol

The Hyper Text Coffee Pot Control Protocol (HTCPCP) is a facetious communications protocol for controlling, monitoring, and diagnosing coffee pots. It is specified in RFC 2324, published on 1 April 1998 as an April Fools' Day RFC, as part of an April Fools prank. An extension is published as RFC 7168 on 1 April 2014 to support brewing teas, which is also an April Fools' Day RFC. Protocol RFC 2324 was written by Larry Masinter, who describes it as a satire, saying "This has a serious purpose – it identifies many of the ways in which HTTP has been extended inappropriately." The wordi...

Is the Banach Tarski's 'paradox' related to Cantor's bijection between lines and areas/volumes ?

Balarka Sen

Q : Why did Bourbaki stopped writing books?
A : Lang was only one person.

Hippalectryon

what is $||A||$ for $A\in\mathbb{R}^n$ ?

Balarka Sen

@Hippalectryon What is $||\cdots||$?

Hippalectryon

11:42

Is it $\sqrt{\sum\limits_{k\in A}k^2}$ ?

@BalarkaSen The norm i think ... not sure

Balarka Sen

Maybe. Euclidean norm, you mean?

Hippalectryon

In a correction of an exercise I have (with some conditions before)

A is a square matrice and $X\in\mathbb{R}^n$

Balarka Sen

Is X also a matrix?

ah.

Then it's Euclidean norm.

Hippalectryon

That is .. ?

Balarka Sen

square root of sum of squares blah

Hippalectryon

11:47

Ok ty

Also math.stackexchange.com/questions/831124/… :)

usukidoll

math.stackexchange.com/questions/831624/induction-and-absolute barf

Hippalectryon

@usukidoll lol ?

usukidoll

what kind of question is that?

does it even have a question?

JMK

@usukidoll looks like the OP must learn some typesetting

Chris's sis

@robjohn have you tried this one? I plan to compute it now. $$\sum_{n=1}^{\infty} (-1)^{n+1 }\frac{1}{n} \left( 1+ \frac{1}{2^2}+ \cdots+\frac{ 1}{ n^2}\right)= \zeta(3)-\frac{\log(2)}{2}\zeta(2)$$

usukidoll

11:54

I actually went apes*** on someone because the op was taking an exam and posting his test paper on here

it was in another language

anyway, anyone on here know partial differential equations?

JMK

@usukidoll That's an irony, I was about to ask the same thing

@usukidoll What is the problem?

usukidoll

well I was just practicing but I think my outline book is making errors yo

Transcript for

$Mathematics$ Mathematics