Mathematics - 2012-01-12 (page 1 of 3)

Srivatsan

12:00 AM

Better. I got 123 points today. Roughly half yours (if I do some division mistakes), I see.

Asaf Karagila

I honestly WTF'd at that last comment.

QED

I don't really understand the notion of polynomial

Asaf Karagila

Polynomials were first defined by Karp in order define what is NP-complete.

@Srivatsan Thank you, I suppose ;-)

Srivatsan

@AsafKaragila I guess I should thank you for the answer. =)

You're welcome, of course.

Asaf Karagila

:-)

I wonder, how did you get to that answer recently anyway?

QED

12:15 AM

For example 3|n^3 + 2n + 3

but if you consider "n^3 + 2n + 3" as a polynomial in Z/3Z it's NOT equal to 0

despite the fact that it equals 0 for every n

Srivatsan

@AsafKaragila Mostly browsing around. That answer caught my attention because that funny comment had come up in our conversation before. (In fact, I might have first gone there for Dylan's answer: when I first reached there, I didn't know it was that question.)

QED

Why are polynomials something different than a subset of functions?

that's what I don't get

Srivatsan

Polynomials are the best. =) No clue why.

QED

what's the benefit of this definition

Srivatsan

Definition of polynomials?

Asaf Karagila

12:19 AM

@Srivatsan I see. Oh well. If you have any follow up questions do let me know.

Srivatsan

- sure, thanks for the offer.

Asaf Karagila

Mark as offensive, so it gets deleted.

QED

why not just wait for the post to get deleted

Asaf Karagila

Because it would get deleted faster if people are active about it.

If something has like 5 or so flags of spam/offensive then it gets deleted automatically.

QED

you're talking about the comment only, though?

Asaf Karagila

12:23 AM

No, it was an offensive "answer".

QED

but there's no way to mark an answer as offensive

oh do you mean "it is not welcome in our community"

Asaf Karagila

Flag it under "it is not welcomed in our community".

QED

yeah, I did that

Srivatsan

I did that too.

Asaf Karagila

I guessed that his friends were trolling through his user, as he said, but little do I care. If you cannot trust your friends you shouldn't leave your user open.

QED

12:28 AM

Why does JM keep writing answers in comments

shoudl post it as an answer so it can be accepted!

Asaf Karagila

Either way, I am going to sleep. I hope the world ends and I don't have to wake up and go to that never ending algebra class! :-)

Srivatsan

@AsafKaragila I hope you get uninterrupted dreams of being forced to sit through never-ending algebra classes.

Sleep tight.

Srivatsan

12:44 AM

What do we do now? math.stackexchange.com/q/98076/13425

QED

nothing?

Srivatsan

Well, either rollback or delete. Something has to be done, but I don't know what's the best route. So I will leave it at that.

QED

let the moderators deal with it

it's been flagged

Srivatsan

Yes, that's my stand as well.

Thanks for flagging - if you did.

QED

don't know if you were interested in it particularly, but I asked the polynomial question and Qiaochu raised an interesting point

Srivatsan

12:48 AM

Yes, I did see it. I had a similar point in mind actually. It feels nicer if Qiaochu writes it though. =)

QED

basically even if distinct polynomials coincide sometimes, they give give distinct functions over a big enough field extension

Srivatsan

In general, it seems like a good idea to separate out the function view and formal view in abstract algebra -- here's a different case of that: math.stackexchange.com/questions/29432/….

QED

That reminds me of the theorem about power series

they converge only in a circle

that was a bit surprise

Srivatsan

@QED "In any case the basic ideas are quite simple if you merely take off your analyst hat and, instead, put on your algebraist or combinatorist hat." -- That sentence is nice. It still does not explain why we define formal polynomials that way though.

QED

yeah

Mike Spivey

1:04 AM

@Srivatsan: It looks like we're starting a conversation in the comments, so I thought I would move it to chat.

Srivatsan

I was thinking we were more like ending our conversation. :)

@MikeSpivey Hi, nice to see you after a long time.

Mike Spivey

Maybe that's true. :) I suppose, since we've agreed that there won't be a bounty, that we probably should delete the comments under the answer.

Srivatsan

So: I guess it's going to be a wee bit awkward for one for us. So I am ok with your not wanting my bounty. Are we agreed on that?

@MikeSpivey Right. We could do that now, just a second.

Mike Spivey

@Srivatsan Yes, I've been busier the past couple of months and haven't been on chat much.

I haven't been on the main site that much the past few weeks, either.

Srivatsan

@MikeSpivey Ah, ok. Busy is also good, I suppose.

BTW, next step: Can you edit out the bounty announcement from the question? Or should I do it?

Mike Spivey

1:08 AM

@Srivatsan Sure, I will edit it out now.

Srivatsan

Thanks. It did feel a little strange for me -- silently pulling down an offer. :=)

Mike Spivey

@Srivatsan I hope I didn't run over you too much there. I just felt a little bad about you giving up 300 rep for me answering my own question. :)

Srivatsan

No, not a problem. =)

Finally: I will leave a comment later saying there was a bounty here. I have to think up something for that.

Ok, I guess that does it then =).

Your solution looks nice, by the way. I have only skimmed through it. It will take me several passes for me to get through it.

Mike Spivey

@Srivatsan Thanks. I really did think about that problem for a long time, and it's satisfying to have found an answer I'm happy with.

By the way, thank you for all you did to promote my question. :)

Srivatsan

@MikeSpivey You are welcome, of course.

Points are to give back, right?

That leaves me with some spare points then. I'll see what to spend it on. =)

So what has been keeping you busy these days?

Mike Spivey

1:20 AM

@Srivatsan Various things... working on some of my own problems, family visiting, general holiday busy-ness.

Srivatsan

Is this you in the picture?

Mike Spivey

@Srivatsan No, I was assigned that picture at random when I signed up for WordPress.

Srivatsan

What? That's a random picture. Quite interesting.

Mike Spivey

Yes, I liked it so I haven't bothered to change it.

Oh, my wife and kids just arrived home so I have to go. Catch you later!

Srivatsan

See you later. Bye!

QED

1:27 AM

It's funny that the linear recurrences can define polynomials and exponentials, but there's nothing in between

Srivatsan

When you say it like that, it is indeed funny. "Between polynomials and exponentials" is a fascinating topic.

This is sort-of relevant: en.wikipedia.org/wiki/Grigorchuk_group, although I don't really understand it.

QED

Now that's weird!

"The group G is finitely generated but not finitely presentable"

that makes a bit of sense

robjohn

2:05 AM

It seems I have missed Mike Spivey. It has been a while since I have seen him here.

Victor

@robjohn - After a week of meditation, i still can't find a reliable or clear intuitive reference for a proof of multivariable changing of variables formula. Please enlight a way for me to get it, at least give me the most important concept during the proof

robjohn

@Victor I have tried to do that, but without a knowledge of matrices and determinants, which you said you did not have, I can't really see a way to do what you want.

Victor

@robjohn - Isn't that you could just expand it? Maybe for 4 variable case?

robjohn

I don't understand what you mean by "expand it". Expand what?

Victor

@robjohn - The determinant...

especially the Jocobian

robjohn

2:14 AM

@Victor Sure, I could expand the determinant, but that wouldn't give you any more of an intuition as to what is going on. You really need to learn about linear algebra (especially matrices and determinants) before tackling multivariable calculus.

Of course the determinant of a $4\times4$ matrix is a sum of $24$ $4^{\rm{th}}$ degree terms. more confusing that enlightening, IMO.

Victor

@robjohn - I think i am able to enpand it, but why does some terms are negetive(the most improtant component)?

So, how does mathmatician knows that that formula is correct?

QED

@Victor, Can you formally state the "change of variables" theorem you want to prove?

robjohn

@Victor That has to do with the fact that when looked at as an n-linear form of the columns (or rows) of a matrix the determinant is an alternating form.

Victor

@robjohn - If i already knows the intuition for 2 and 3 variables case, does that help?

robjohn

@Victor One proves the formula from the fact that the determinant is an alternating n-linear function of the rows (or columns) of a matrix and that the determinant of the identity matrix is 1.

Victor

2:24 AM

@robjohn - How do you show it? i maybe able to read it

robjohn

@Victor you get that intuition from visualizing the geometry of the situation. The geometry of 4 dimensions is harder to visualize.

Victor

@robjohn - Different dimentional case is very different?

Example: 4 and 5?

QED

@Victor - do you understand my question?

robjohn

@Victor do you remember the last time we spoke about this? You were having trouble because you were not able to generalize your intuition from 3 dimensions to 4.

@Victor The abstraction of matrices allows us to view all dimensions pretty much on even footing.

QED

@Victor ?

Victor

2:29 AM

@robjohn - i think if i could overcome the 4 dimension case, i may understand all cases, also, what is the prelimitaries on definition of 4 dimension other than linear algebra

robjohn

@Victor do you understand that any vector in $\mathbb{R}^n$ can be written as a linear combination of the standard basis vectors?

Victor

@QED - I think i am not sophisticate enough to make a formal statement

@robjohn - I know 3 variables case have 3 vectors from a point and so on...

QED

@Victor, You need to be able to do that if you want to understand and prove it

Victor

@QED - Can you write that for the 4 variable cases?

robjohn

@Victor a standard basis vector looks like $\begin{bmatrix}0\\0\\\vdots\\0\\1\\0\\\vdots\\0\\0\end{bmatrix}$. all $0$s, except a single $1$.

QED

2:31 AM

I don't know what it is you want to prove

robjohn

@Victor: I have to leave for a while. I will see later if you are around and we can perhaps do more to prove the formula for a determinant. Think on what I have said so far.

Victor

@robjohn - So how do you make a definition(geometrically) or lemma to prove the 4D case

@robjohn - I wonder why a math professor is that busy at night

@robjohn - Thanks

Victor

3:00 AM

@robjohn - Are you there?

t.b.

3:26 AM

@Srivatsan you might like that, actually. Chapter VIII of de la Harpe's book is entirely devoted to it. The group has an easy enough description as a group acting on trees and also in terms of finite state automata. This handbook chapter is also quite nice.

Srivatsan

@tb oh! Thanks.

If I ask you how you know this stuff, would it make sense? [I am asking this because many a time, I have been told that such a question doesn't make sense and is just funny. I think they are all wrong.]

Is this something all or almost all math people would have seen?

t.b.

@Srivatsan Long story... I was very interested in geometric group theory when I was working on my diploma and at the beginning of my thesis. The Grigorchuk group is one of the major sources of counterexamples for many natural conjectures. The first conference I attended was in honor of Grigorchuk's 50th birthday.

Srivatsan

@tb Oh ok.

t.b.

@Srivatsan I guess not, but people interested in infinite group theory, probably yes.

Srivatsan

Just clarifying.

t.b.

3:40 AM

The question on growth is a very nice one. I think. Given a finitely generated group, what is the asymptotic growth rate of the size of $n$-balls in the group.

Srivatsan

Do you know [about] Gromov's theorem? I don't know (and not looking for an explanation now: that would be at best a bed-time story to me right now). I have seen Terry write about it a lot, hence I ask.

t.b.

Sure. (March 31)

(that was before the proof via harmonic analysis by Kleiner)

Srivatsan

Heh

t.b.

(and before Tao's blog, too :))

Srivatsan

@tb In which year was this conference(?)?

t.b.

3:45 AM

2005

It was more like a reading seminar.

QED

I don't understand this comment math.stackexchange.com/questions/98345/…

"You can think of polynomials as functions" didn't I just give a counter-example?

Srivatsan

@tb Seeing this comment made me realise I subconsciously revealed how I am secretly on a first-name basis with Terry Tao. =) In fact, it's so secret that he doesn't know this yet...

robjohn

@Victor First of all, I am not currently a math professor. Second of all, why should a math professor not have a life at night?

Srivatsan

@robjohn Not "life at night"; "busy at night". How can a math professor not read messages properly? =)

robjohn

@Srivatsan :-p

Perhaps I was being too harsh. Perhaps not :-)

Srivatsan

3:50 AM

I felt like saying something as much. I decided that would be out of place for me.

robjohn

I don't think it would have been out of place.

t.b.

@QED I think that's the way people thought about polynomials before the algebraic revolution around the beginning of the 20th century. It looks like a valid point.

Srivatsan

Clear answer, that one was.

robjohn

@tb what is virtual nilpotency?

t.b.

@robjohn a group is virtually (P) if it has a finite index subgroup having property (P)

robjohn

3:54 AM

okay, sounds reasonable.

t.b.

that's why I'm not particularly interested in finite groups -- they're virtually trivial!

(standard joke, I admit)

Srivatsan

So (to myself, mostly): it's a group satisfying (P) + a little bit extra.

QED

haha

robjohn

@Srivatsan well, $1/n^{\rm{th}}$ of a group has the property

Srivatsan

@robjohn Are you saying I should've said a group satisfying (P) times a little bit extra? =)

robjohn

3:57 AM

so the little bit extra could be quite a bit more than the part satisfying (P)

@Srivatsan that would be better :-)

Srivatsan

[I was going to give a lame example to show that I understand, but nevermind.]

t.b.

What? four people upvoted this nonsense?

Korgan Rivera

hey all.

Oh great wise ones on high, please link me to a simple explanation of where the exponential decay function comes from.

Srivatsan

Wait, what!? The great wise ones are on high? =)

Korgan Rivera

They're on something.

Srivatsan

4:07 AM

@KorganRivera Exponential decay function? What is that?

Korgan Rivera

@Srivatsan, you know, along the lines of ce^-(ƛ t)

Srivatsan

Oh, that's just the exponential function. I guess decay refers to the fact the function monotone decreases to zero as $t \to \infty$.

Do you know about radioactive decay?

Korgan Rivera

That's a big question :)

But yes, I understand how the function can be used to measure the amount of a material with a given half-life after some amount of time.

I just want to know how this function is derived from scratch.

Srivatsan

First: have you seen this Wikipedia article on radioactive decay?

Korgan Rivera

I read the one on exponential decay, let me check through this one..

The explanation has a "let's rearrange the terms" step, and they use dy/dt notation, and their step is to 'multiply' by dt. I'm trying to avoid the whole infinitesimal thing.

Srivatsan

4:15 AM

So, -from a physical point of view- the defining characteristic of the exponential function is the "first order growth or decay". That is, if $X$ is the amount of some substance, and if the rate of change of $X$ is directly proportional to how much amount is currently present (i.e., the derivative of $X$ is proportional to $X$ itself), then you end up with such a differential equation; the solution happens to the exponential function.

Korgan Rivera

I guess I can work it out from there, but if anyone has a link to something else I'd like to see it.

Srivatsan

It's not really infinitesimal thing: it's a differential equation. You can solve it rigorously as well.

Here's how to do it:

$$ \frac{dX}{dt} = - \lambda t .$$

Rearrange it to: $$ \frac{dX}{dt} + \lambda t = 0. $$

Now the magic step: multiply by $e^{\lambda t}$. It is the "integrating factor" for the differential equation. Then an application of the product rule will simplify it to:

$$
\frac{d}{dt} \left( e^{\lambda t} X\right) = 0.
$$

Is everything ok till now? Can you go from here?

Korgan Rivera

I thought the first step would have been $\frac{dX}{dt}=-ƛX$

Srivatsan

I don't understand; isn't that what I wrote?

Korgan Rivera

no you said it's equal to -ƛt

btw, mathjax isn't rendering on this page, should it?

Srivatsan

4:21 AM

Ah, sorry. Both my first and second equations are wrong. They should be $\lambda X$, as you say.

Korgan Rivera

ok.

Srivatsan

See this page for MathJax support in chat.

Korgan Rivera

Ok this is what I understand so far. So, the proportion of our substance reduces as a proportion of its initial 'weight' over time. So I can say dX/dt = -ƛX. After that, I'm stumped. I suppose I find a solution for X by integrating?

thx for the link.

Srivatsan

@KorganRivera I wrote down the solution - except for the last step. Of course, correct the first two equations.

Korgan Rivera

k thx.

Yeah that didn't help, but thanks :) 'Magic step' threw me off. I'm sure it's not that complicated, I'll find a solution somewhere.

Srivatsan

4:32 AM

Do you want me to try this once more - this time splitting the magic step into two?

Korgan Rivera

If you have the patience, I would appreciate it. Can you start from where my understanding ends?

Srivatsan

Your understanding ends within you; how would I know where it ends? =)

Korgan Rivera

Because I told you.

4th post before this one.

..of mine.

Srivatsan

OK. The third step is to multiply by the integrating factor of this differential equation, namely $e^{\lambda t}$.

Korgan Rivera

Ok, but where does that come from?

Srivatsan

4:35 AM

OK: do you know about the exponential function? Or is all this an indirect way of asking us to explain that?

[I'm not scolding you or anything; I don't know if my tone suggests that. I'm not sure what you know and what you wish to know through chat.]

Korgan Rivera

No no I know. You're trying to help. I'm not sure if I know enough intuitively about the exponential function to help me here. I thought I did.

Let's carry on with this path to see if it clicks for me.

so I have $\frac{dX}{dt}e^{\lambda t} = -\lambda X e^{\lambda t}$

Srivatsan

@KorganRivera Most of the times, the discomfort is related to "how do I manipulate this expression or solve this function", not what that function "is" or what the expression "means".

If your trouble is the latter, I cannot help you that much; since to me also, some bits appear out of the blue.

Korgan Rivera

I understand. :)

Let's carry on.

Srivatsan

Ok. Let's agree that multiplying by $e^{\lambda t}$ is the right thing to do. I can give some more intuition - but later. Not now.

Korgan Rivera

Ok, now what?

Srivatsan

4:43 AM

So: $$ e^{\lambda t} \frac{dX}{dt} + \lambda X e^{\lambda t} = 0 .$$

Korgan Rivera

ok

Srivatsan

Now it's time to recognise patterns: in fact, this is why we multiplied by $e^{\lambda t}$ in the first place.

Korgan Rivera

Right now, I'm seeing a massive loss of spirits. What do you see? :)

Srivatsan

Let's call $e^{\lambda t}= Y$ temporarily. Then what is $Y'$?

Korgan Rivera

$\lambda e^{\lambda t}$

Srivatsan

4:45 AM

I take it that you mean $\lambda e^{\lambda t}$. So: in terms of this new function, we can rewrite the equation as: $Y \frac{dX}{dt} + X \frac{dY}{dt} = 0$.

Korgan Rivera

yes

Srivatsan

Does this look better?

Korgan Rivera

oh yes. It does.

Srivatsan

@KorganRivera Do you want to take over and complete the solution?

Korgan Rivera

I want you to drag me there like a corpse.

Srivatsan

4:48 AM

@KorganRivera You don't even want to try?

Korgan Rivera

uh.

Srivatsan

Well, that is not quite what I meant. Because both the terms involve X.

Korgan Rivera

Yeah I just saw that :(

robjohn

Integrating factor

Srivatsan

@KorganRivera HINT: The expression $Y \frac{dX}{dt} + X \frac{dY}{dt}$ should remind you of some specific thing.

Korgan Rivera

4:50 AM

oh it's the product rule.

Srivatsan

@KorganRivera So, what is the next step - after applying the product rule (actually, you should apply the product rule in "reverse" =))?

Korgan Rivera

Well at this point I have $\frac{d}{dt}(e^{\lambda t}\cdot x)=0$

Srivatsan

Yes. So, the derivative of a particular function is zero. What could that function be?

Korgan Rivera

I feel I ought to be very embarrassed at how slow my progress is right now. Just to assure you, I'm well aware of it :)

That function would be some constant.

Srivatsan

The function $e^{\lambda t} X(t)$ is a constant; call this $C$. I made it $X(t)$ for emphasis.

Korgan Rivera

4:56 AM

Ok.

Srivatsan

$e^{\lambda t}X(t) = C \implies X(t) = C e^{- \lambda t}$.

Korgan Rivera

Ok that makes sense.

Thank you I appreciate you doing that. I'll look at this tomorrow morning and wonder what was so difficult about it :)

Sivaraman

@Srivatsan Can I get a couple of minutes? :)

Srivatsan

Where are you?

Sivaraman

GHC - office

Korgan Rivera

5:04 AM

@Srivasan: that multiplying by $e^{\lambda t}$ is a very clever step. I wouldn't have figured that out in 50 years.

actually I would have, but I wouldn't have figured it out tonight.

chat.stackexchange.com aka the skeletal trebuchet.

..always throwing me bones, thanks folks please tip your waitresses, g'night!

robjohn

5:47 AM

@Korgan I don't know if that will ping @Srivatsan

N3buchadnezzar

8:25 AM

-hi =)

Skullpatrol

@N3buchadnezzar Hi, it's so quiet in here ... that you can hear a pin drop before it hits the floor!!!

(3 hours later...)

N3buchadnezzar

I was just dropping by to ask a simple question, before going to my lectures ^^

Skullpatrol

@N3buchadnezzar How simple?

N3buchadnezzar

I have a parametric curve defined by the expression

$$x=t \sin(t) \: , \: y = t^3 $$

My goal is to find where this parametric curve fails to be smooth.

My first idea was that the function fails to be smooth, where the derivative is undefined. eg

$$ a = \frac{dy}{dx} = \frac{3t^2}{\sin(t) + t \cos(t)}$$

Now "obviously" this expression is undefined when t=0. But it is also undefined for an infinite number of other points.

My question, is why are these points discarted? By my graph, it looks like the only point where the function is not smooth is when t=0. But my calculations says otherwise.

Skullpatrol

@N3buchadnezzar Maybe you can "ping" one of the professors ... if you set the denominator equal to 0 and solve for t ... what do you get?

N3buchadnezzar

8:34 AM

Its unsolvable in terms of an explicit expression

Skullpatrol

Have you tried polar coordinates?

Kannappan Sampath

Hello @Matt

Matt

And *inside* research analysts it's too dark to see general convergence in topological spaces.

Hello Kannappan.

I'm a 4k user!!! How did that happen!?

t.b.

You explained that $\omega + 1$ has a last element... :)

Matt

: )

Rahul Narain

9:12 AM

@N3buchadnezzar The curve can fail to be smooth only if both dx/dt and dy/dt are zero.

Daniil

9:46 AM

Hi

robjohn

@RahulNarain or if $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ or $\dfrac{\mathrm{d}y}{\mathrm{d}t}$ are not smooth

heh, it appears I've missed Rahul.

Asaf Karagila

10:17 AM

Hm.

Daniil

10:47 AM

almost started with my coursework :D

robjohn

11:41 AM

@Daniil started? don't get ahead of yourself :-)

Skullpatrol

@robjohn I found the Algebra Bible

robjohn

@Skullpatrol and what would that be?

Skullpatrol

@robjohn Algebra Bible

Kannappan Sampath

@Skullpatrol You mean Dummit and Foote

Skullpatrol

@KannappanSampath Brown and Dolciani et al

Kannappan Sampath

11:46 AM

never heard of it :(

Is it a modern algebra text

Skullpatrol

@KannappanSampath The link says "Algebra II Bible"

Kannappan Sampath

@Skullpatrol My head is composed of mud! that's some classical algebra text.

But, good that I came to know of it!

Skullpatrol

@robjohn Today I came across a book in the library entitled "Is God a Mathematician" and now I find an Algebra Bible, do you think that this means that mathematics is a religion to some people Rob?

robjohn

@Skullpatrol It has proofs of what it says and relies not on faith. I think that disqualifies it from being a religion.

Skullpatrol

@robjohn What did you think of the "Algebra II Bible?"

Daniil

12:08 PM

blackboardsinporn.blogspot.com/2010/09/5.html haha

(not very safe for work)

robjohn

@Skullpatrol Are there only 6 chapters?

Skullpatrol

@robjohn I think that is just the study guide link.

robjohn

@Skullpatrol yeah, but I only saw six chapters. How many chapters does "Algebra and Trigonometry Structure and Method" have, I wonder?

Skullpatrol

@robjohn 16

@robjohn Have you read "Is God a Mathematician?"

robjohn

12:27 PM

@Skullpatrol No

@Daniil This one at least has some non-trivial (and correct) math.

Daniil

It also looks like something I would watch

robjohn

@Daniil anything to learn, eh?

Skullpatrol

12:43 PM

@Daniil Do you prefer "formulae" or"formulas?"

Daniil

12:57 PM

Latter.

Skullpatrol

@Daniil Latter.

Skullpatrol

1:23 PM

@robjohn My 12 year-old nephew asked me "How can I be sure I'm right when I solve an equation?"

robjohn

@Skullpatrol depends on the equation. Usually, you plug the values back into the equation and see if it works.

Skullpatrol

@robjohn That is what I told him, but he said "why is that true?"

robjohn

That doesn't make sense. You are looking for values to make an equation true. That means if you plug those values into the equation, it is true. How more basically can you verify it?

Solve $x^2+y^2=25$ and $x+y=7$. $x=3$ and $y=4$ is a solution, and we check that by $3^2+4^2=25$ and $3+4=7$. There may be other solutions, but at least that is a solution.

Skullpatrol

@robjohn He wanted to know why it works or what makes the method work.

robjohn

Why what works? plugging values in to check? I don't understand what is confusing about that.

Skullpatrol

1:34 PM

Yes, plugging values into check.

robjohn

you want to solve $x^2-3x+2=0$. Saying that $x=1$ is a solution, is simply saying that $1^2-3\times1+2=0$.

Skullpatrol

I had to tell him "If
the original equation
is true for
some
value of x,

robjohn

Find out what he thinks solving an equation means. If he means something else, I would like to know.

Skullpatrol

then
the real number properties guarantee
the final equation
is also true for
that
value of x,

robjohn

That confuses me. The equation is true for some value of $x$ when you substitute that value of $x$ into the equation and the equation is true. $x$ is simply a placeholder for some value.

Skullpatrol

1:40 PM

where the final equation is a transformation of the original equation.

robjohn

Perhaps he is confused about the role of a variable.

Skullpatrol

Perhaps.

robjohn

Oh, well, I was assuming you were plugging the values back into the original equation.

Skullpatrol

I was.

robjohn

Then where is the confusion?

Skullpatrol

1:43 PM

but he wanted to know where I got the value to plug in?

robjohn

what is being asked when I ask for a solution of $x^2-3x+2=0$?

@Skullpatrol Didn't you show him how you solved the equation? Plugging back in is just the check.

Skullpatrol

I said I solved the equation. We only worked on linear equations.

I said to him

robjohn

The check should be the test of the correctness of an answer.

Skullpatrol

I agree.

But kids like to ask why, why, why,...

robjohn

Is he confused about why if $3x+2=11$ then $3x=11-2$?

Skullpatrol

1:48 PM

no

but when we say x = 3, and sub in 3

he asks why

robjohn

Is he asking why we check?

Skullpatrol

no, how do we know if the 3 is right?

robjohn

Because we checked it by plugging in

Skullpatrol

I said because it satisfies the original equation

The transformations work both ways

with the exception of multiplying by 0

robjohn

It doesn't matter that the transformations work both ways. You have plugged the values in to the original equation, and that shows that they solve the equation.

Skullpatrol

1:56 PM

True.

robjohn

If he is asking why we solve linear equations as we do, then it is necessary that the operations are reversible.

Skullpatrol

Yes, I think that he is asking that.

robjohn

But if I want to solve $3x+2=11$, subtracting $2$ from both sides says that $3x=9$ dividing both sides by $3$ says that $x=3$. So we have said that if $3x+2=11$ then $x=3$. Without reversible operations, we have not said that if $x=3$ then $3x+2=11$, but that is why we check (even if the operations are reversible).

Matt

I wonder what's going on with the moderators. I have two pending flags on two separate posts one of which is waiting since yesterday.

robjohn

In the case of $x^2-3x+2=0$, we usually complete the square to get $x^2-3x+\tfrac94=\tfrac14$ so that $(x-\tfrac32)^2=\tfrac14$

This is different. There is no single reversible operation that will give us $x$.

We can factor $(x-\tfrac32)^2-(\tfrac12)^2=(x-2)(x-1)=0$

Then we have that either $x=2$ or $x=1$, but we can't know what $x$ was; $2$ or $3$. We just know it is one or the other.

But we still check the answer by plugging in.

I have to go for a while. BBL

Skullpatrol

2:10 PM

@robjohn In your first example Rob, if x being 3 is the value that satisfies the original equation then plugging it back in will give a true statement (provided no errors have been made), then have we said if x = 3, then 3x + 2 = 11?

robjohn

Plugging in shows that if $x=3$ then $3x+2=11$

The derivation of the solution shows that if $3x+2=11$, then $x=3$

Skullpatrol

@robjohn But you said we have not said that if x=3 then 3x+2=11, but that is why we check (even if the operations are reversible).

@robjohn So why we check is to show the converse is true. That is why we check.

Asaf Karagila

2:31 PM

@Matt It sometimes takes some time to think about the flags, if they are borderline cases. Also I recall Mariano saying that sometimes he's not sure whether or not to accept the flag and leaves it to the other moderators. I'm certain it is the same with the others as well.

@Dylan Hi.

Dylan Moreland

Hey Asaf.

How are the sets?

Asaf Karagila

What's up?

Dylan Moreland

Oh very little. Teaching in an hour. Reading more Deligne.

Asaf Karagila

What do you teach?

Dylan Moreland

It's this linear algebra/calculus of several variables hybrid course.

Asaf Karagila

2:43 PM

Who's in the audience?

Dylan Moreland

It's awkward for the students, because they don't really like being precise but any course involving the abstract definition of a vector space is going to require that somehow.

First-year students.

Undergraduates.

Asaf Karagila

Math? Engineering?

Dylan Moreland

"Science"

Broadly speaking.

I don't think they're math majors, by and large.

I'm going to grab some coffee. Back in a bit.

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$Mathematics$ Mathematics