@usukidoll do you know spanish?

no sorry

mabye you would have liked to do my induction homework

o-o

depends on the problem

Prove that $\sum_{k=1}^n\frac{1}{\sqrt n}\geq\sqrt n$

yeh, like that

that one should be induction

so you sould use the fact t is true for n to prove it is true for n+1

Are you gonna give it a try?

awwwwwwwwww man x.x

I've read about this problem

that one is the second problem in my homework

the first one is trivial

basis should be 1

and then omg the P(n+1) ... killer

ol

lol

yes, usually the base for induction is easy

$\sum_{k=1}^n\frac{1}{\sqrt (n+1)}\geq\sqrt (n+1)$

:/

@user4140 Dat der is a triviality. Think you mean $\sum_{k=1}^n\frac1{\sqrt k}$ on the left.

Oh lol

yes

@Mike lol, I honestly totally forgot how to do it, and I can't figure it out either.

anyways, give it a try @usukidoll

:/

I can do it for you and give you number 3 in the list, but they get harder...

can't use the binomial theorem on this that's for sure wrong format

there are 10 problems

it's not one of those fjlfjlkdas is divisible by whatever either

no

I know give me an example on how to tackle them

I'll do this one and you do the next.

and then maybe I can use that guideline to prove these bad boys... or try to

@Mike I looked it up in Isaacs' algebra. It's elementary, but not a triviality.

Hint: Pick $y\in G\setminus\langle x\rangle$ such that $|y|$ is minimal.

Oh, and you have to induct on $|G|$.

So do you want me to give solution?

I found it.

ugh how would I know?

Ok, first we prove for n=1. It is true since $1\geq1$

yes

Now we shall assume that for a given n we have $\sum_{k=1}^n\frac{1}{\sqrt k}\geq \sqrt n$

yup

@Karl Induction was unexpected. Thanks, I'll give that a shot.

We have to prove that $\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\geq \sqrt {n+1}$

-.-

So if we prove $\frac{1}{n+1}\geq \sqrt{n+1}-\sqrt{n}$ we are done

why $\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\geq \sqrt {n+1}$ not $\sum_{k=1}^n\frac{1}{\sqrt (n+1)}\geq\sqrt (n+1)$

too fast man

The problem is to prove $\sum_{k=1}^{n}\frac{1}{\sqrt k}\geq \sqrt n$

for all positive integer values of $n$. I had written it wrong the first time, as karl commented

ok?

@user4140 It took me forever to prove the induction step. facepalm

ok

We already proved base ok?

in other words we proved the statement is true for n=1.

So now we shall prove that if the statement is true for n then it is true for n+1.

In this case we will prove if $\sum_{k=1}^{n}\frac{1}{\sqrt k}\geq \sqrt n$ then $\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\geq \sqrt {n+1}$

Ok???

doll?

@usukidoll you there?

yes

we on the same page?

oh you've changed the limits on the sum

yes

so maybe that's why the n became n+1

that's exactly why it became n+1. If the limit would have been w then the number in the root would have been w.

Because we want to prove that the sum of the multiplicative inverses of the roots of the first n posotive integers is larger than the square root of w.

ok?

anyways, what we are gonna do is assume it's true for n and then using that we are going to prove it is true for n+1.

yup

so we have $\sum_{k=1}^{n}\frac{1}{\sqrt k}\geq \sqrt n$

Now what would we have to add to both sides to get $\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\geq \sqrt {n+1}$??

well clearly $\frac{1}{\sqrt{n+1}}$ to the left and $\sqrt{n+1}-\sqrt{n}$ to the other side right?

So uf we where able to prove the thing we are adding in the left is larger to the thing we are adding to the right then we would be done right?

since if something is larger than another and we increment the large one by more than the small one then the larger will still be the large one right?

yeah... hey what is going on with your latex?

So now we are going to prove $\frac{1}{\sqrt{n+1}}\geq \sqrt{n+1}-\sqrt{n}$ directly

$\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\geq \sqrt {n+1}$ so we're adding $\frac{1}{\sqrt{n+1}}$ where the $\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\geq \sqrt {n+1}$ is and $\sqrt{n+1}-\sqrt{n}$ to that $\sqrt(n+1)$

Right

To prove that we are going to rationalize the equation

rationalize the numerator?

$\frac{1}{\sqrt{n+1}}\geq \sqrt{n+1}-\sqrt{n}$ if and only if $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}}\geq (n+1)-n$ if and only if $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}}\geq 1$

Ok?

nvm I don't think it's called rationalixing

I thhink it's called conjugating or something.

Do you follow?

>:/

WHAT! there is such a thing as rationalizing the denominator and the numerator

I just multiplied both sides by $\sqrt{n+1}+\sqrt{n}$

yeah

and I got a difference of squares in the right side.

So you understood the proof?

I did problem 3 while explaining problem 2, so you have to do problem 4, it's not too hard I think.

-.-

sort of I wish there was a youtube video of it

You have to prove that if the vertices of a convex n-gon are colored with at least 3 colors such that no two consecutive vertices have the same color then you can divide it into triangles using only straight cuts between vertices of different colors.

This one is with strong-induction

so you have to use the fact it works for all values smaller than n+1, and then prove it also works for n.

Give it a go!

NEVER! >:O

why not?

why can't it be easy like $2+4+6+...+2n=n(n+1)$

It's actually really easy.

Here is the proof:

like for basis on that let n = 1 2 = 2

then the P(k) so that's 2k=k(k+1)

Take the convex n-gon. Make any cut between two vertices of different color.

P(k+1) = 2(k+1) = k+1(k+2)

$k^2+3k+2$ is what I need to achieve

lol, you told me you would help me in my HW

ok, There is one like that in the homework.

@Karl I'm just not seeing how to reduce it to a simpler case. I can't just mod out willy-nilly.

viola ... reached my goal line

nice

I need to go to sleep, but You will help me with my HW tomorrow, mark my words.

nyah nyah I got school tomorrow :P

and I only know that one and the $5^n-6^n$ is divisible by 4 problems ... that's just an example btw

Like, let's say my element of minimal order is $|y|=4$ but that $y^2 \in \langle x \rangle$. I can't just get rid of $\langle y\rangle$ and hope to use the direct factor thing there.

true or false there are so many varieties when it comes to induction besides what I posted above

@usukidoll Here's one you might find interesting $$\sum_{k=1}^n\frac1{\sqrt{k}+\sqrt{k-1}}$$

omg

hides

@usukidoll try rationalizing the denominator, just as you were doing above.

blah can't do it well on latex

I know that I have to multiply both sides by the denominator but i have to change the $+$ t o a $-$

@usukidoll You mean multiply by $\dfrac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k}-\sqrt{k-1}}$?

yeah

then the denominator would be $k-(k-1)$

but $k-k+1$ so that's $1$

@usukidoll yes

whew my algebra is still intact after all these years

it's the analysis parts that ARE DRIVING ME NUTS! WHy introduce such a thing so late !!!!!!

I can do the easy proofs but anything hard and I'm lost fast... even some pdf files won't work

so it either has to be a very good book or a very good video for me to understand

@usukidoll So you now have $$\sum_{k=1}^n\left(\sqrt{k}-\sqrt{k-1}\right)$$

does that look familiar?

yes it does

I missed whatever you removed :(

Oh, I made a silly error, as I was confused between additive and multiplicative notation. I am fixing.

My hero!

I should say my thing is abelian, so. This isn't true I imagine for general groups.

@usukidoll can you compute the sum? If not, try writing a few terms out.

hmmm....can someone help me with this diff eq problem?

@JakeShellman We cannot possibly know that answer until we see the problem!

@JakeShellman no one will know until you ask :-)

@JakeShellman first order or second order?

sorry..:) here is the problem:

y' = y^2 + ay + 1

compute the sum? @robjohn

The question asks to determine equilibrium solutions based on the parameter a

$y'=y^2+ay+1$

@usukidoll $\displaystyle\sum_{k=1}^n\left(\sqrt{k}-\sqrt{k-1}\right)$

we hadn't finished the problem

Wouldn't you just set the autonomous equation equal to zero, then use quadratic formula to solve for y?

ARGH! I do know that once you determine the equilibrium, it's 0 a horizontal line on the slope field graph

@JakeShellman any range given for $a$?

stupid question how do I compute the sum...I need my memory refreshed. if I just plug in $k=1$ I got 0 man

ummm no I just errr converted the question into latex

nope

@robjohn nope

@usukidoll try writing out a few terms...

So....I think you just generally solve using quadratic equation right?

like?

k k+1

k-1

$k^2$?!

y = (-a +/- sqrt(a^2 - 4)/2

these are the equilibria right?

O_O

@JakeShellman to factor the denominator and then use partial fractions to integrate

well, you don't need to solve the ODE to figure out equilibria right?

@usukidoll the sum... write out the first few terms...

:O

@Mike Derp, you first have to decompose $G$ into its $p$-components. Then work with each $p$-group alone.

of $\displaystyle\sum_{k=1}^n\left(\sqrt{k}-\sqrt{k-1}\right)$

k... ummm k-1uhhhh

@robjohn you don't need to solve the ODE to figure out equilibria right?

@Karl Can't. No structure theorem.

most of the time no...just factor @JakeShellman

@JakeShellman you can usuallly do without

@Mike By $p$-component, I mean the set of all elements of order $p^n$ for various $n$.

Obviously a direct summand.

Oh, gotcha.

Haaha, was about to ask before you edited.

@usukidoll @robjohn so then, just use quadratic formula to find the equilibria, no?

@usukidoll what is $(\sqrt1-\sqrt0)+(\sqrt2-\sqrt1)+(\sqrt3-\sqrt2)$

most of the time you can just factor and find x = 1 2

OH! that's what it is

$(\sqrt1-\sqrt0)+(\sqrt2-\sqrt1)+(\sqrt3-\sqrt2) =(1+(\sqrt2-1)+(\sqrt3-\sqrt2)$

@JakeShellman that will give you the points where $y'=0$, yes

which are by definition equilibrium solutions right?

@robjohn

@JakeShellman yes

@Mike Do you see how you can't have $y^2\in\langle x\rangle$ in an abelian $2$-group?

@JakeShellman do you know how to tell if the equilibria are stable?

and then you plug in other n's greater than -1 less than -1

ACK!~!!!!!!!!!!!!! I just remembered something.. don't u have to take the derivative to find out if it's stable unstable or semistable

@robjohn So then you could look at the parameter a and since $y' = y^2 + ay + 1$, you could say that if a = 2 or a = -2, there is only one equilibrium solution (plus or minus 1) and if a is between -2 and 2, then there are no equilibria...and if a is less than -2 or greater than 2, there are two equilibria

._.

if I remembered correctly the derivative greater than 0 is unstable.. less than 0 stable and 0 semistable...my prof didn't test the class on this

@Karl Teah, I can reconstruct the proof from your tip.

but I have read about it

@usukidoll stability is determined by the value of y' for values of t less than or greater than the equilibria. So, if y' tends to be positive and negative (toward the equilibrium) it is stable, but if they point away it is unstable (in terms of the phase line)...if they both point in the same direction, it is termed semistable

That's also the hardest one since the rest allow me the structure theorem, so I can just beat them roseate.

to death

ah now I see it...

like I said prof didn't make a test question or provided homework on this subject

same for the mass problems

math.stackexchange.com/questions/670464/… downvote the bottomanswer like the wind

@Karl What does 'finite intersections' mean? That you only consider the interseciton between finite sets or that you only consider intersections which are themselves finite?

@Kevin Intersection of a finite number of closed sets

@Mike Thanks

Err

Finite number of open sets, sorry

(Intersection of arbitrarily many closed sets is still closed)

@Mike I was about to ask you to torment me back as I was stuck on a problem, but I solved it while typing it. :(

@Karl Haha, I wish I had that skill.

bites @KarlKronenfeld

It's "Let $M\subseteq\mathbb R^n$ be an embedded submanifold. Show that $M$ has a tabular neighborhood $U$ with the following property: for each $y\in U, r(y)$ is the unique point in $M$ closest to $y$, where $r:U\to M$ is the [canonical retraction]." As far as I can tell, we are fine using any tabular neighborhood, right?

Oh, I guess we have to worry about $M$ "bending back" too closely.

That is easy to deal with though...

Tubular, you mean?

I can't help, though. I'm doing differential (topology/geometry) next quarter.

yes

@Mike Taking advantage of the relationship between a tubular neighborhood and the normal bundle, it's basically an application of the fact that tangents of $M$ at $x$ are orthogonal to $(y-x)$ in $\mathbb R^n$, when $x$ is a closest point of $M$ to $y$.

If it works, it works.

Heya @PedroTamaroff

how do I make a pleasant $||\cdot||$?

where the bars aren't spread like so

\lVert\cdot\rVert

lolwut

$\|$

or just \|

would you be my tutor?

works for me

that's @Karl, not @usukidoll

I don't tutor unless I get paid

@usukidoll I would be the least dedicated internet tutor ever.

I wasn't asking you @Mike

awwwwww @KarlKronenfeld :(

omg @mike sharing is caring though

@usukidoll And that's why you should share the wealth with me

>:/

t(-_-t)

@Mike What is your charge though?

@Sawarnik $1 higher than your price range

pfffttt

@Sawarnik The key is I'm not in it for the money ;)

@Mike Then?

@Sawarnik I'm not in it for the money... I'm not in it without the money... so I'm not in it at all :)

@Mike Perhaps in with the money would be better choice.

Maybe, maybe.

going to bed night

@usukidoll What is your time zone?

@AlexYoucis Isn't it past your bedtime, Shadow?

@Mike Aren't you in the same timezone as me, brohamine?

I'm gonna have to steal brohamine

That's patented.

Just like Shadow.

Get off my lingo turf.

You've been swaggerjacked.

That sounds offensive.

I'm offended.

It's a great word. You're free to swaggerjack the word.

Never. I am original--through and through.

The Shadow seems pretty swaggerjacked to me.

You're swaggerjacking a long history of villains.

I think you're mistaking who stole whose handle.

I expect you use a new language, for authenticity purposes, @AlexYoucis.

@KarlKronenfeld I do--I just speak english amongst other non- PeetsyDoodles speakers.

That's a terrible name for a language.

You should honestly be ashamed of yourself.

@AlexYoucis Oh, what if I became a PeetsyDoodle speaker? What is "hi"?

@Mike Me and Ribet speak it to each other when we are discussing our most recent gourmet conquest--on yelp.

@KarlKronenfeld Hi=hi.

Sheesh.

You people.

You people? Are you presuming to reduce an entire culture to "you people"? You disgust me.

I'm pretty sure the culture is two people, tops.

@Mike 12, if you must know.

Every joke I think of makes it sound like I'm OK with marginalizing minorities, so you win this time.

As if that wasn't my strategy.

You just have to accept the fact that I am superior to you on many banter levels.

I'm attempting to grade homeworks but it's hard to grade fairly when every person gets the question wrong in increasingly bizarre ways.

What course?

(also, you've just implicitly admitted my superiority)

$1+1=e$...$1+1=\infty$...$1+1=\frac{\pi}{\emptyset}$...

@AlexYoucis I've only allowed you to make yourself look silly by gloating over something so trivial.

Also, PDE.

@Mike If you consider the conversational subjugation of others trivial, then you were never a worthy opponent to begin with.

Also, gross.

There's a reason I'm chatting here instead of just doing it.

PDEs can be really nice, I just have no faith that what you're grading is.

No, of course it's not. It's solving Laplace on a square with certain boundary conditions.

Hey, neither do the students

I bet they aren't as witty as me though, yeah?

One of them likes to tell me I lead a sad life, but the rest haven't realized they're allowed to insult me on their homeworks yet

No comments from him this week though

I hope you just write "lol" on there.

Well, I did the same thing to him when I was grading his number theory homeworks, so I've had it coming.

Sounds like he doesn't have much more of a life than you do.

We're roughly tied.

Well, he got 100% this week, but he doesn't get a smiley face.

You just have the caged tongue of a university employee.

Haha, that'll show him--little shit.

Think fast:

do you math blog?

That might be the harshest thing anyone's ever said to me.

No.

Never.

:/

I math blog.

I don't see the point of it unless someone else will benefit, and I haven't found anything that I think I have an insight someone else hasn't yet.

Maybe I have, who knows.

I do it mostly for me

lol

PS, can I show you something stupid that always makes me laugh

Eh, that's what my notes are for

Go or it

Iloleverytime

I 3/4's lold