Mathematics - 2016-11-11 (page 7 of 7)

saturatedexpo

18:10

$\leq$ is transitive or?

Astyx

yes

over $\Bbb R$

saturatedexpo

18:24

if someone wants to look over this if the formatting seems ok?

(inductions, so the language doesnt matter much)

Astyx

Mm ?

saturatedexpo

ufile.io/c8418

Astyx

Your "Induktionsschritt" seems a bit weird

saturatedexpo

@Astyx hello how are you? :)

Astyx

Fine and you ?

saturatedexpo

18:30

yes me too :)

Mithlesh Upadhyay

How to find the sum of series 2+3/2+1+5/8+....

saturatedexpo

you mean from the first or second excersice?

Astyx

first

saturatedexpo

@MithleshUpadhyay is this fibunacci?

@Astyx it's a mess, but it's right. I might wewrite this with factoring out (j+1)

Astyx

I would write $${1\over6}j(j+1)(2j+1) + (j+1)^2 = (j+1)({1\over6}j(2j+1) + (j+1))$$ and simplify the RHS until I get ${1\over6}(j+1)(j+2)(2j+2)$

saturatedexpo

18:32

instead of multiplieing everything out

Astyx

Yes it is write

But stating : "we want to find" and then go from there seems strange to me

saturatedexpo

mmh, i first rearranged the left side, then wanted to show that this equals to the RHS

Astyx

@MithleshUpadhyay What is the general term of your series ?

teadawg1337

@MithleshUpadhyay So the n-th term is $\frac{n+1}{2^n}$, correct?

saturatedexpo

and with equalities at least you can even devide by (j+1) if you really wanted i guess (altho it was not adviced to me!)

Mithlesh Upadhyay

18:34

@saturatedexpo , how, please explain? I don't know. fib=1,1,2,3,5,8,13,...but there is divide operator?

teadawg1337

I don't see what this sequence has to do with the Fibonacci numbers

Semiclassical

@mithlesh first, you need to find the pattern. do you see what it is?

saturatedexpo

@MithleshUpadhyay it's not very clear how the series looks like to me :/

do you have the explicit form?

Semiclassical

(hint: divide each term in the sequence by 2)

saturatedexpo

@Astyx ah now i understand your "prob"

teadawg1337

18:37

It's too late for me to delete that post, so the n-th term is given above

Astyx

@saturatedexpo For the second exercise, the $\implies$ sign does not mean what you think it means

saturatedexpo

you mean i should only take the LHS and try to form it into the RHS?

teadawg1337

That's my bad

Astyx

What you did at the first exercise seems formally correct to me

But a little odd

saturatedexpo

well \implies is pretty self explanatory ? :(

Astyx

18:38

Then again, maybe that's just me

saturatedexpo

$\implies$

Astyx

The issue with what you wrote is that $A\implies B$ does not mean that $B$ is true

Mithlesh Upadhyay

@teadawg1337 , , how you find the n-th term is $\frac{n+1}{2^n}$ ?

Astyx

What you wrote is true but it is not sufficient to prove the inductive step

saturatedexpo

ah!

mmh

maybe instead of the arrow: "it follows..."?

Astyx

18:40

Yes

Semiclassical

@MithleshUpadhyay that doesn't quite match the sequence you provided: this one goes (starting from n=0) 1, 1, 3/4, 4/8, 5/16,...

which is a factor of two off from the series you gave

teadawg1337

Sorry

$\frac{n+2}{2^n}$

Semiclassical

yeah

Mithlesh Upadhyay

@Semiclassical , yes , my series seems wrong? but its a 12th grade final exam problem.

teadawg1337

Then we can't help you at all

Semiclassical

18:43

eh, the series you wrote out is perfectly valid and doable.

saturatedexpo

@Astyx i think, if i have the free time i wewrite the first proof. i think you are right. The LHS should be viewed seperate from the RHS. Otherwise it's easy to get errors.

Semiclassical

but what exactly are you getting stuck on? finding the nth term, or going from there to summing it?

Mithlesh Upadhyay

@teadawg1337 , that OK. :)

teadawg1337

I personally object to helping someone with a final exam that they're taking

Semiclassical

oh, this is a final exam problem you're doing yourself right now?

then yeah, nope.

Astyx

18:45

Why ?

If I may ask

saturatedexpo

to whom "why"?

Mithlesh Upadhyay

@Semiclassical , its 12:15am and there is no exam rightnow :)

Astyx

Why not help someone for their final exam

teadawg1337

@MithleshUpadhyay So is this a problem you found online?

@Astyx Hypothetically speaking, it would be a breach of honesty in the test-taking environment. It's technically cheating

Mithlesh Upadhyay

@teadawg1337 , thanks for your help. Just, I'm doing assignment for newspapers; LOL

Astyx

18:48

Do you mean for the actual exam that they are taking ?

Semiclassical

for newspapers?

Astyx

Or like last minute help

teadawg1337

Last minute help is perfectly fine. Asking for help in the middle of an exam, however, is not

Astyx

Oh yes, this I completely agree with

Semiclassical

i guess i'd also object if it were a take-home exam problem

teadawg1337

18:49

That as well

Mithlesh Upadhyay

@teadawg1337 , thank;'

@Semiclassical thanks

Semiclassical

if one only wants to sum the series, try multiplying it by two term-by-term and then then take the difference

saturatedexpo

well, final exams are retarded anyways. My girlfriend had a E-grade, then took a oral exam and got a B.

(in math)

Mithlesh Upadhyay

@saturatedexpo , WOW

Semiclassical

if you do the subtraction in one way, you'll get back your original series; but if you do it another way, you'll get a geometric series plus a constant.

Mithlesh Upadhyay

18:52

@Semiclassical thanks again

Semiclassical

np

saturatedexpo

@Semiclassical you mean something like a term paper?

"if it were a take-home exam problem"

Semiclassical

nah

in upper division courses you occasionally see exams where consisting of a problem set that you take home and then return a few days later

so basically a really tough homework set.

saturatedexpo

and are those also really hard even if you know all the stuff involved? i.e. it takes much time even if you understand it?

Semiclassical

googling, i get this as an example: physics.umd.edu/courses/Phys402/cohen06/takehome.pdf

saturatedexpo

19:03

wow impossible in one week haha (from my pov)

Semiclassical

hence why it 'works' as a take home exam

saturatedexpo

show for an ordered field: $$|x-z|\leq |x-y|+|y-z|$$ what principles do i have to know to be able to show this?

Tobias Kildetoft

@saturatedexpo Basically just the definitions

saturatedexpo

so write the absolutes out and eventually get there? or make cases?

Tobias Kildetoft

"write out" and "make cases" is the same for absolute value

saturatedexpo

19:10

oh ok

Mithlesh Upadhyay

@Semiclassical , I am still troubling because answer should be 6, and symbolab gives =1/2+4?

symbolab.com/solver/step-by-step/…

Balarka Sen

rewatched a film by one of my favorite directors

Mithlesh Upadhyay

@teadawg1337 , I stuck

Semiclassical

where are you seeing 1/2 instead of 2?

plus, all i see from symbolab is that the series converges (not what it converges to)

saturatedexpo

@MithleshUpadhyay if you showed that it converges you can set $$a_n=a_{n+1}$$

Balarka Sen

19:19

@TedShifrin Hey. You're back home yet?

Semiclassical

what perhaps you need to focus on is showing that $\sum_{n=0}^\infty n/2^n = 2$

Mithlesh Upadhyay

@Semiclassical yes

?

Ted Shifrin

Hi @Balarka — nope, sitting on the Berkeley campus. Turns out no classes today ...

Semiclassical

not sure what the ? is for

Balarka Sen

@Ted Ah.

Semiclassical

19:21

what symbolab tells you is that $\sum_{n=0}^\infty \frac{n+2}{2^n} = \sum_{n=0}^\infty \frac{2}{2^n}+\sum_{n=0}^\infty \frac{n}{2^n}=4+\sum_{n=0}^\infty \frac{n}{2^n}$

uh, no. that's nto what I was saying

saturatedexpo

ok

Semiclassical

what i was saying is that i didn't follow where he was confused

Balarka Sen

I think someone I know is at Berkeley right now on a geometric group theory workshop.

Ted Shifrin

Hi @Semiclassic @teadawg

Semiclassical

hi @ted

saturatedexpo

19:23

@TedShifrin are you the Ted all are talking about? :D

Semiclassical

@ted I think I've figured out that the 'method' i came up with is just the brute force method i've used before, just presented in a different way

sort've disappointing :/

Mithlesh Upadhyay

@Semiclassical I mean how to $\sum_{n=0}^\infty n/2^n = 2$?

Semiclassical

yeah, that's the hard part.

Ted Shifrin

There are a few people around, but the math building is locked up as tight as a prison. So I can't get in. What a way to treat alumni!

saturatedexpo

make an upper bound

Semiclassical

19:24

if you only want to show that said series is finite (which is all you need for convergence) then you don't need to show that it actually equals 2

Ted Shifrin

@MithleshUpadhyay: Write down a Taylor series and differentiate/integrate it.

saturatedexpo

make a "lazy" guess on the highest value this can approach to, then check with epsilon method

Balarka Sen

@TedShifrin Yikes.

Ted Shifrin

Then evaluate at $x=1/2$.

Astyx

Or use Cauchy Product

saturatedexpo

19:26

yyyy

Ted Shifrin

@saturatedexpo: Doubtful you can do that.

Semiclassical

the shortcut way of doing it is to first expand it out: $$S=0+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\cdots $$

Mithlesh Upadhyay

And, as always answer is 42? I confused :)

Semiclassical

noting the factors of 2 in the denominator, we multiply it by 2 to shift them all:

Ted Shifrin

There is a trick for this one, like the trick for the geometric series.

Semiclassical

19:27

$$2S = 1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\cdots $$

Ted Shifrin

Which is what Semiclassic is saying.

Semiclassical

yeah.

if I now subtract those two series term by term, what do I get?

saturatedexpo

is this legit?

Balarka Sen

Yes, all of that is perfectly legit.

Astyx

$$\left(\sum_{k=0}^{+\infty}{1\over 2^k}\right)\left(\sum_{k=0}^{+\infty}{1\over 2^k}\right) = \sum_{n=0}^{+\infty}\sum_{k=0}^n {1\over2^k}{1\over2^{n-k}} = \sum_{k=0}^{+\infty} {n+1\over 2^n}$$

That is legit because the series are absolutely convergent right ?

ie the famillies are summable (is that the term in english ?)

Semiclassical

19:30

well, the radius of convergence of 1/(1-x) is 1, and we're working at x=1/2

so everything's fine, yeah.

there's a lot of ways to do this problem, it should be said

the way @Ted said initially revolves around the differentiation map giving $x^n \mapsto n x^{n-1}$

saturatedexpo

can one think of a graphical proof that a sum converges?

(any sum, just never saw something like that)

Astyx

Well ploting the partial sums ?

It's not really a proof

But it can give you an idea

saturatedexpo

ah so you basicly want to show that: theres an asymtote?

and that the values dont hop between lets say 0 and 1

Astyx

For instance

Semiclassical

two proofs-without-words of the geometric series: albany.edu/~bd445/…

upper left picture here is another: usamts.org/About/U_Gallery.php

and more here: artofproblemsolving.com/wiki/…

Mithlesh Upadhyay

19:40

, I really confused, so i posted it, please try to explain
http://math.stackexchange.com/questions/2009669/find-the-value-of-sum-n-0-infty-left-leftn2-right-2n-right

Semiclassical

that title is a bit misleading: you ask about $\sum_{n=0}^\infty (n+2)/2^n$ in the title, but in the post itself you make clear that you're only confused about $\sum_{n=0}^\infty n/2^n=2$

and, well, that's a series that i'm sure has shown up on MSE before.

@MithleshUpadhyay you might compare with this similar problem: math.stackexchange.com/q/1343721/137524

Mithlesh Upadhyay

That I have understood and other not, so an explanation will be good

I also think, i need to sleep.

saturatedexpo

@Astyx "it follows" means $\iff$ or?

Semiclassical

"A follows from B" means $B\implies A$

Astyx

There is no symbol for "it follows"

And using words is not something you should try to avoid

saturatedexpo

19:46

yeah just asking^^

i mean if a statement follows from another, then those two statements are equivalent or?

probably bad phrasing

We have a>0. It follows that $a^2>0$

Semiclassical

If I'm talking on this chat, I must've logged in. Hence, from my messages being here, it follows that I'm logged in right now.

Can you conclude from that fact that, if I'm logged in, I'm talking on chat?

saturatedexpo

no

Semiclassical

Correct.

So A follows from B is an implication, not an equivalence.

Astyx

Well you could formally write $A \land (A\implies B) \vdash B$

Or something similar, I'm not an expert

But that's just bad

Semiclassical

I forget what \vdash means

saturatedexpo

19:49

ah, that what i wondered about

yeah doesnt seem nice

Astyx

I'm not sure, I might have mistook this symbol for an other

saturatedexpo

but just wondered what the heck "it follows" means :p

we got A verified

and we know A->B

that means B can be concluded?

Semiclassical

If A is true and it's true that A implies B, then B is true.

which is known in logic as modus ponens

Astyx

Yes

saturatedexpo

"white holez is black holez of anti matter!!" i fail to see the fail :s

saturatedexpo

20:21

is the math olympiade every year new problems?

Astyx

Of course

user243301

20:35

@AndersonFelipeViveiros This is only to do feebback about your question. A thing important in the live of vikings was the navigation in all climatic conditions. Thus you can search what's about it. I don't know references. Search references in internet also in spanish, or other languages.

Balarka Sen

21:31

what does the Poisson bracket do, geometrically? I don't quite get why it should be a vector field (I can prove it).

by Poisson bracket, I mean Poisson bracket of vector fields, defined by $[X, Y](f) = X(Y(f)) - Y(X(f))$, where $X(f)$ and the like means directional derivative of $f$ along $X$.

arctic tern

that's the Lie bracket right?

SteamyRoot

Yeah, it's Lie

Mike Miller

yes, not Poisson

arctic tern

it measures the rate of change of Y(f) under the one-parameter group of diffeomorphisms exp(tX) or somesuch

Balarka Sen

if you say so. I don't really know what the Lie bracket is. that's what Milnor calls it

Mike Miller

21:37

@Balarka it's the infinitesimal diffeomorphism given by flowing a little bit on X, then a little on Y, then backwards on X, then backwards on Y

arctic tern

I never liked the parallelogram explanation as much as the conjugation one

Balarka Sen

@MikeMiller ah, so by that you mean that's what flowing along [X, Y] does I suppose?

Mike Miller

depends on whether you like thinking of vector fields as derivations or infinitesimal diffeomorphisms

I am quite suspicious of the claim that Milnor calls that the Poisson bracket

SteamyRoot

Yeah

It must be a typo

Balarka Sen

@Steamy It calls it Poisson at least twice

Astyx

21:41

Poisson bracket

In mathematics and classical mechanics, the Poisson bracket is an important binary operation in Hamiltonian mechanics, playing a central role in Hamilton's equations of motion, which govern the time evolution of a Hamiltonian dynamical system. The Poisson bracket also distinguishes a certain class of coordinate transformations, called canonical transformations, which map canonical coordinate systems into canonical coordinate systems. A "canonical coordinate system" consists of canonical position and momentum variables (below symbolized by qi and pi, respectively) that satisfy canonical Poisson...

SteamyRoot

Then he was confused, most likely

The Poisson bracket is something different (though of course it has similar properties

math.tecnico.ulisboa.pt/~acannas/Books/lsg.pdf

Check page 108 of the pdf (116/225 in pdf readers)

Balarka Sen

dunno, I'd be rather careful about saying Milnor's confused. but sure.

no point fighting over terminology

SteamyRoot

Well, they are very closely related and have similar properties, so it doesn't matter much

Either way, perhaps mathoverflow.net/questions/127792/… is of use to you?

Balarka Sen

thanks; but Mike already told me the answer I wanted.

Alessandro

good evening

Mike Miller

21:54

@Balarka In any case this is the first and only time I've ever seen someone call that the Poisson bracket.

Astyx

It's called the Poisson Bracket in Classical mechanics for what I know

Balarka Sen

interesting. will call it the Lie bracket from now on if that's the proper word.

Mike Miller

@Astyx Yes, but that's a different thing.

Astyx

That's what I figured from the discussion you just had :)

WDUK

23:27

Could some one help explain how the solution set is worked out from the graph of the inequality x^2 + x - 2 < 0. I got (-inf,-2] U [1, inf). But it seems it is actually (-2,1) i don't understand why

given the line goes on forever why does my book state it is actually closed at those numbers not infinite =/

TheGreatDuck

hello

Is there anyone here that could help explain what I am trying to say to a user? It's a pretty complicated concept and I think they're having a hard time understanding my question.

It seems you have many difficulties explaining what you want. You never talked of "a final operator" — user1952009 21 mins ago

thanks

oops

0

Q: How can I define alternate indefinite integrals that give different results rigorously using this method?

I wish to rigorously define alternate different integrals where an arbitrary set of functions is what all potential integrals of some kind vary by (by "vary by" I mean that the difference of any two integrals is an element of this function set), and where the alternate integral treats functions o...

Stupid Man

In the formula for sum of infinite arithmetico geometric series, $S_n = \frac{a}{(1-r)} + \frac{rd}{(1-r^2)}$ what is $a$? Is the initial value of the series of the initial value of the AP part of the series?

Ted Shifrin

@StupidMan: Yes, the series is $a+ar+ar^2+\dots$.

Not sure about your formulas there.

Stupid Man

23:44

@Ted Shifrin I mean this series en.wikipedia.org/wiki/Arithmetico-geometric_sequence

At the end of that page there's a formula for the infinite AGP series. I wanted to know what the variable $a$ represents in that formula. Is it the initial value of the AP part or the initial value of the AGP series. Thanks.

Ted Shifrin

I've never seen that before, but they show you what $a$ is. Write out the first few terms.

TheGreatDuck

anyone have any idea on how I could make my question better?

I tried editing it.

I'm just not sure how clear it can become.

Ted Shifrin

@TheGreatDuck: I'm not going to look on my phone, but offhand you need your function set to consist of solutions of some particular differential equation.

TheGreatDuck

the lack of rigorous notation, terminology and other things makes it difficult to express. I'm trying my hardest but it seems like it won't get answered anytime soon (the only person commenting seems to have trouble understanding), which is quite sad as having a way to express this for other people to understand would be quite beneficial to me.

"I'm not going to" huh? I asked if you had any ideas on how to improve it?

"but offhand you need your function set to consist of solutions of some particular differential equation." how so?

are differential equations a way of altering the integral?

Ted Shifrin

Patience, it's hard for me to type. Don't be an ass.

TheGreatDuck

23:51

I am being patient.

Ted Shifrin

I can't do chat on my phone. Sorry.

TheGreatDuck

ok

by alter the integral do you mean like (for instance) constructing an integral that treats the floor function as a constant?

It just seems bizarre that diff eq would actually work as a method.

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$Mathematics$ Mathematics