Mathematics

2:00 PM

@TheArtist You said it as if I knew yours. ;) Chris's sister the artist isn't good enough? :D

@DanielFischer I seem to recall Hardy providing some sufficient conditions, but of course I can't find the paper or book at the moment. The conditions I stated are most likely a misinterpretation of what he stated.

The Artist

@Chris'ssistheartist no but I am trying to quote you :P

Chris's sis the artist

lollll :-)

The Artist

@Chris'ssistheartist fine, that will do .

Chris's sis the artist

@RandomVariable I never asked you if you attend integration for many years. No idea about sos either (excepting the period in which I joined MSE - he was already in place with great answers).

Daniel Fischer

2:02 PM

@RandomVariable Possible. But I have no idea where you should look.

The Artist

@Chris'ssistheartist A daring journey on a $\int$

Random Variable

@Chris'ssistheartist I didn't even take a calculus course until college.

The Artist

@Chris'ssistheartist google told me that "escapading" is not the verb of "escapades" , so it does not really work.

Chris's sis the artist

@TheArtist It's not easy to find a good name for a book. :-)

@RandomVariable I see.

Agawa001

@Chris'ssistheartist really?, making a book is easier than choosing a proper name ?

Chris's sis the artist

2:07 PM

@Agawa001 I only said it's not easy to find a good name for a book. :-)

The Artist

@Chris'ssistheartist totally agree with you.

Agawa001

i wonder how much not easy to write a book that the name isnt even easy to find

The Artist

@Chris'ssistheartist would not say its easier than writing a book. Writing a book is very difficult.

Chris's sis the artist

@TheArtist It's very difficult but at the same time it's an amazing journey, you learn so much while doing that. :-)

Agawa001

if i made a book, i would find a suitable name for it

i would name it for example, "what you dont expect to read in other books" lol

2

Chris's sis the artist

2:11 PM

:-)))))))))))

The Artist

@Chris'ssistheartist This came to my idea :P Title : "Finding True Love" . Centered below the text is the sign \int ♥ dx

Chris's sis the artist

@TheArtist :-)))))))))) People would think it's about some novel. ;)

The Artist

@Chris'ssistheartist yes correct. And its important to have what the book is about in the title

@Chris'ssistheartist : "Living Inside an Integral"

@Chris'ssistheartist sharing treasures and things you found in life by living with integrals ^^^ ;)

Chris's sis the artist

@TheArtist lollllllll, nooooooooooo :-)))))))))))

Back in 30-60 min.

The Artist

@Chris'ssistheartist lol

Agawa001

2:17 PM

working offline is time saving, take this advice from a programmer :)

see ya

The Artist

@Agawa001 language?

Agawa001

maternal language is C

i know java and javascript and vb

as well

and you cant be a complete programmer if you dont know some asm basics

so i do

The Artist

@Agawa001 is that your profession too? :)

Agawa001

i guess so, i spend lot of time programming stuff

The Artist

@Agawa001 cool ^_^

Agawa001

2:25 PM

its really diificult, sometimes you cant satisfy people's needs, sometimes you repeat something done dozen times (deja vu), so it turns annoying

but the main thing is fun in its globality

M.S.E

@Agawa001 it must be. I failed my "Introduction to Programming for engineers" course

@Agawa001 and we did C and Matlab.

Agawa001

lol, have you tried harder ?

The Artist

@Agawa001 I have done HTML , super easy lol

and CSS

Agawa001

there s nothing easier than programming, its like you tell the machine to do something, and it does nothing apart what you order it

:)

M.S.E

@Agawa001 i put that aside now. Will have to repeat it next year so will try more then, but yes I did word hard.

Agawa001

2:28 PM

just take it as simple

The Artist

I think its an easy subject for some people and not so good for other people.

user159870

The sum of roots of the equation $z^{2014}=i$ is equal to what? How can we know?

Agawa001

its super easy to understand a dumb-machine, what i find harder is understanding social life :(

The Artist

@user159870 De Movire's?

Daniel Fischer

@user159870 Generally, when you have a polynomial, how is the sum of the zeros connected to the coefficients?

M.S.E

2:30 PM

@Agawa001 what is so hard about social life?

Daniel Fischer

@M.S.E Somehow, people tend to be involved. And people are difficult.

M.S.E

@DanielFischer involved in?

Agawa001

@M.S.E well, not all people think the way you do, and thats lil bit conflictual

Daniel Fischer

@M.S.E In social life.

M.S.E

@Agawa001 that is why you talk to as many people and find the ones that share common interest and take it to the next level to build a relationship.

@Agawa001 Isn't it a game of probability? The rest (majority) of the people you say "see you around" and never talk again.

user159870

2:34 PM

@DanielFischer You mean this? the sum of the roots is $\frac{-a_{n-1} }{a_n}$, $a_n$ is the leading coefficient and $a_0$ the constant.

Daniel Fischer

@user159870 Yes.

Now, what are your coefficients when you want the solutions of $z^n = a$?

Agawa001

@M.S.E its complicated, more events can make things more complicated, not just one or two variants, its more thorough

M.S.E

@DanielFischer but how are they difficult? :) look at this chat, no one is difficult here.

Daniel Fischer

@M.S.E You are new here, right?

M.S.E

@DanielFischer lol yeah but still so far, in hours of a few weeks I have been here :P

Agawa001

2:38 PM

@M.S.E wait to see questions of topology, you would probably change your mind

user159870

@DanielFischer $a_n=1, a_{n-1}=0$.

Yes?

M.S.E

@Agawa001 okay :P

lopata

if 1+1 =2 why 1-1 ≠ -2 ?

M.S.E

@Agawa001 but similar thinking people gives rise to less complication rate like 50% will have no problem.

Daniel Fischer

@user159870 Yes. (For $n > 1$.)

lopata

2:41 PM

How can PI exists if circle aren't provable?

M.S.E

@lopata Because + means going right 1 step in the number line, and - means going 1 step left on the number line

Agawa001

@M.S.E i can find 70% of people who think in my way just in internet (usually people are more plugged into material)

lopata

How do you prove PI exists?

Remember me

@Balarka I have to prove that $G \cong Z_p$ where G is an abelian simple group. Now since G is simple the only subgroups it will have will be 1 and G. Similarly for $Z_p$. So if I define the map $\phi$ to be the one which takes $\phi(1) = 1$. I got till this much. How should I proceed after this

user159870

@DanielFischer So there are no roots. Yes?

M.S.E

2:43 PM

@lopata a(sign + or -) b=c , means a is the initial position, sign is the direction to move, b is the number of steps to take, and c is the final landing position

lopata

lol

Remember me

@lopata $\pi$ is defined to be the ratio of circumference to the diameter

Hello@TedShifrin

lopata

I mean, how do you prove cricumference exists if perfect circles don't exists?

M.S.E

@Agawa001 that is true :)

Remember me

@Ted A question :
How do I prove that a group $G$ of even order is isomorphic to $Z_p$ . What happens if $G$ is of odd order

lopata

2:46 PM

@Rememberme you have to prove the injonction of abelian group G in subgoup Z

Remember me

@lopata Many undefined terms there :
1) What do you mean by perfect circles?
2) What do you mean when you say circumference exists

@lopata injonction ?

Balarka Sen

@Rememberme take an element in your abelian simple group and look at the subgroup generated by a single element from your group.

recall that a subgroup of an abelian group is normal

Daniel Fischer

@user159870 No. Counting multiplicities, every polynomial of degree $n$ has exactly $n$ zeros in $\mathbb{C}$. That the sum of the zeros is $0$ doesn't mean there are no zeros.

lopata

@Rememberme I meant Injection

Injective function

In mathematics, an injective function or injection or one-to-one function is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain. In other words, every element of the function's codomain is the image of at most one element of its domain. The term one-to-one function must not be confused with one-to-one correspondence (aka bijective function), which uniquely maps all elements in both domain and codomain to each other, (see figures). Occasionally, an injective function from X to Y is denoted f: X ↣ Y, using an arrow with a barbed...

Random Variable

@DanielFischer It's way too much to post here, but here's a link to the paper if you're interested: play.google.com/books/…

user159870

2:52 PM

@DanielFischer So can we say that the sum of roots of $z^{2014}=i$ is $\frac{a_{2013}}{a_{2014}}=0$?

Remember me

Well isomorphic means there exists a bijection $\phi$ such that $\phi(xy)=\phi(x)\phi(y)$. Now I have to find a bijection which has this property @lopata

Daniel Fischer

@user159870 Yes.

lopata

I can't see latex

Agawa001

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