What is the best way to practice mental maths

@MohamedAhmedNabil Mental maths?

like working out 17*15 in your head

@MohamedAhmedNabil Why'd you want to do that? I mean, it just comes with time I guess.

I usually do 100+50+70+35, for example

255

LOL

@MohamedAhmedNabil Memorizing the times tables is the fastest way to get the answer.

As Peter has shown the "best" way is to be able to break it down into smaller problems.

17*15 = (10+7)*15 = 10*15 + 7*15 = 150 + 70 + 35 = 150 + 105 = 255.

ok thanks

What about mapping/drawing quadratic equations in my head

Oh ok

@WillHunting True dat.

what is the difference between calculus and algebra

@MohamedAhmedNabil Algebra is about mathematical strctures, calculus is in general about limiting processes and analysis.

@WillHunting Am I correct?

@WillHunting Hmm. @MohamedAhmedNabil Don't worry about that somuch then.

@anon You there?

@PeterTamaroff then what is pre calculus

yea

@MohamedAhmedNabil What you need to know to start calculus courses.

@anon I have some doubts about uniform continuity.

@WillHunting No no.

UC in general.

and by doubts you mean questions, I presume?

The definition is that $f$ is UC in $A$ if for every $\epsilon >0$ there exists a $\delta >0$ such that, for every $x,y \in A$; if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon.$

funny innit?

@anon Whatevs.

Now, the idea is that $\delta$ depends only on $\epsilon$; correct?

well, $\delta$ can depend on $f$ and $A$ if you want to be technical

@anon Yes, OK.

So, for example, is $\sin(x^2)$ UC on $\mathbb R_{\geq 0}$? I think not.

@anon Why $0$?

I think the problem is for large $x$, right?

I was thinking sin(1/x), nevermind

@anon =P

but yeah, if you extend out to infinity it won't be UC

because x^2 keeps growing quicker and quicker

@WillHunting ...

nvm, you deleted...

anyone saw this live youtube.com/watch?v=KCQVXEgSbrk&feature=player_embedded?

in fact of $f$ is periodic and $g$ is strictly monotone increasing on (0,inf) with lim g'->inf then $f\circ g$ shouldn't be UC

@WillHunting OK.

assuming f isn't constant anyway

wait I said strictly nevermind

HI @ZhenLin

@WillHunting Hi Will!

skullikins

yo anon

@PeterTamaroff Did you find out what distinction your Prof was making between : and |?

@JohnJunior Yes,

@PeterTamaroff Drum roll....

roooolllling...

The suspense... It is killing me.

He told me $:$ means "which verifies" while $\mid$ means"such that". He said it is merely a literal distinction, not mathematical.

Hm

is programming connected to maths?

@WillHunting Let me translate it with a dicc.

a better word would be satisfy, because it's not like elements are each detectives going out verifying things.

"veryfing" is the word.

@anon It is common use in Spanish.

@anon Sometimes English and Spanish have different word which seem to mean the same, but have different uses.

@PeterTamaroff Is that a Spanish literal distinction? and what do you mean by "literal"?

@WillHunting I will kill you with a popsicle if you do so.

@PeterTamaroff what do you mean by "literal"?

@WillHunting By my popsicle.

@JohnJunior That it is not a matter of maths but of language.

As in literature ???

he meant literal, not literary (having to do with literature)

OH GAWD people come on!

@PeterTamaroff Literally come down from where?

@WillHunting In the literal sense.

What is the literal sense of verify?

To verify the truth.

I bet this room currently has more language discussion than ELU's..

So given these conditions are true = such that = ":" = "|"

(:

The vertical version really caught on. :)

@WillHunting Death by popsicle

@anon I bet this room currently has more language discussion than ELU's $\Huge\text{..}$

$\Huge\bullet\bullet$

@WillHunting What is your real name?

David Webb

INfamous Webb

@anon What is your initial? (Come on, taht at least!)

@WillHunting Wanna know is initial.

He won't tell his name.

@WillHunting ¿?

$\Huge\bullet\bullet\text{anon}\bullet\bullet$

@anon Booooooooooo!

@JohnJunior HAHAH

@WillHunting That's clearly a Thai trans.

a glamour model from Toronto, Canada, it would seem

in other news, google now allows you to do searches with images, not just for images

how long have I been out of the loop on that?

dunno

I think Jordan dropped out.

@anon Yes, my cell has that. I take a pic and it searches stuff.

I should try taking a portrait.

@WillHunting You owe.

@WillHunting If you have any personal reason not to do so, I'm OK with it.

totally helpful

I keep getting empty files though :/ (I have the newest version and third-party cookies enabled, so I can't identify the problemo)

yes, but that's easy to overcome. just click every single option it gives..

youtube

indeed

Nice: pt.scribd.com/doc/70421640/…

I can't recall the last time I've seen this room quiet for 3.999... hours.

hi @JonasTeuwen

@WillHunting Your avatar flies off and comes back a lot.

Hi.

@JohnJunior It certainly is unencumbered by complexity.

@JonasTeuwen Good morning

Morning.

We're going to to stay at the Y-M-C-A.

The original YMCA Music Video from 1978 has 27,794,396 views!

But the music video by Justin Bieber performing Baby has 778,505,348 views!!!

We forgot Rebecca Black. =(

Today is friday!

I prefer black friday ;-D

Perhaps the human race has gotten progressively dumber since Fermat's time.

@peoplepower Why?

@GustavoBandeira Because most people do not take this exit.

@peoplepower Idiocracy.

@peoplepower Be careful with this conception - Focusing illusion may be around.

@GustavoBandeira Yes, only the great achievements from the far past make their way into the present. My earlier remark was a joke rather than a belief of mine.

@peoplepower Yep. =)

Good morning!

hi

hi @hhh

What is the sum of a serie? The limit to which it tends?

If it tends...

Yeah, right :P

Motion, thought of as a word problem, is a paradox.

Umh?

Nope, starting from A and walking till you reach B

@unNaturhal At some point in time don't you have to be half-way between A and B?

@JohnJunior Yeah

@unNaturhal OK, so from that point A' to B don't you have to be half-way between A' and B?

@JohnJunior Yes..

Something like Zenone paradox?

and from that point A'' to B don't you have to be half-way between A'' and B, yes Zeno's paradox...

What are you trying to tell me?

Since there is an infinite number of points between any two points and you have to be at each of them you can never go from A to B :-D

Perfect...

The same theory can be applied to my exam

You have to go from the first theorem on the textbook (A) to the last one (B)

Since there is an infinite number of middle theorem in between, and you have to study each of them, you will never ends your class!

As you said an infinite series may have a finite sum...

This would be correct if the Zeno's paradox was correct.. but it doesn't is..

@JohnJunior This is correct instead

Zeno's paradox is correct as a "word problem."

But not ad a logic problem

Right...

Motion in the "real" world has a time constraint, because the intervals of time get shorter and shorter between the mid points; and this series converges at point B.

Yeah, just like Achille and the turtle.. Finally, he will reach and pass the turtle..

Damned series...

@WillHunting Who?

@WillHunting Why not?

The orange is suposed to "high-light" your messages, no?

@WillHunting Yeah I know.. But I can't understand why this argument (series) is too hard for me ._.

@WillHunting The Cauchy test for series is not necessary? (it's really hard ._.)

A series is called "infinitesimal" when it's limit is equal to 0?

@unNaturhal Don't know but the size of my brain is surely infinitesimal!

@JonasTeuwen In my head, infinitesimal is something really small, and infinite is something really big ($\infty$).. but the Leibniz test says that to use this test the serie has to be "infinitesimal and decreasing"..... ?

@JonasTeuwen What gonna happen now? :P :P

@JayeshBadwaik Hmm, I believe the thing is they put the pictures which are not yet published on another website and then link to it and claim they have nothing to do with it :-).

Hmm.

hi @RichardSullivan

hi everybody

You can never have enough Johns.

We are dense in the set of all the participants

Yes, for any two members in the set of all participants, we can find another John that is between those two members and is also a member of that set ;-)

What is the ordering on the Johns...?

I'll let John answer that.

there is no ordering

we've been all made equals

ok in fact I was just testing the chat

see you next time

bye

@WillHunting Aren't the two definitions equivalent (in the case the space is metrizable, so that the "between" is defined)?

@WillHunting A subset $A$ in a metric space $X$ is dense $\iff$ $\bar{A} = X$ $\iff$ the only open set disjoint from $A$ is $\phi$. There is a full list but this will do for now. Consider there is an open set disjoint from the set of johns. And now consider, there are points $A$ and $B$ which are "johns", so now we can find a point $C$ between the two "johns", which is again a "john". Now, the disjoint set is either between $A$ and $C$ or between $B$ and $C$.

Now by using the definition recursively and continuing, we can see that, the open set must either contain at least one john.

With apologies to all Johns.

@WillHunting I trying to prove the thing of UC.

hi @anon

$$\left| {f\left( x \right)g\left( x \right) - f\left( y \right)g\left( y \right)} \right| \leqslant \left| {g\left( x \right)} \right|\left| {f\left( x \right) - f\left( y \right)} \right| + \left| {f\left( y \right)} \right|\left| {g\left( x \right) - g\left( y \right)} \right|$$

hey

Now, I assumed $f$ or $g$ bounded. But I'm not sure if I need both of them bounded.

If they are both bounded, I can work it out.

Because I get

$$ \leqslant {M_1}\left| {f\left( x \right) - f\left( y \right)} \right| + {M_2}\left| {g\left( x \right) - g\left( y \right)} \right| < {M_1}\frac{\varepsilon }{{2{M_1}}} + {M_2}\frac{\varepsilon }{{2{M_2}}}=\epsilon$$

Right?

@anon I wanted to ask you what you think of adding the word "Askaway" to the title of the room to encourage people to stop saying "Can I ask about ..."?

For $\delta_3=\min(\delta_1,\delta_2)$

@PeterTamaroff Both need to be bounded. $x\sin(x)$ is not uniformly for example.

Only one function is bounded is possible when the domain is bounded though.

@JayeshBadwaik I don't understand how your example applies.

@JayeshBadwaik If the domain is bounded, then the function is necesarily bounded.

room topic changed to Mathematics: Associated with Math.SE; for both general discussion & math questions alike (no tags)

@PeterTamaroff I meant to say that, the condition that only one function is bounded is okay only when the domain is bounded.

@JayeshBadwaik Right.

@JayeshBadwaik But then again, if the domain is bounded, both functions are necesarily bounded.

They are uniformly continuous, which means they are continuous, thus bounded.

So, I meant to say that, if the domain is not bounded, then both functions have to be UC for the product to be UC as my example says.

what does domain mean, precisely? 1/x is not bounded on (0,1) for instance.

@anon But you lost the link?

@JayeshBadwaik OK. Again. The hypothesis is "If $f$ and $g$ are uniformly continuous on $A\subseteq \mathbb R$; and at least one is bounded, then $f\cdot g$ is uniformly continuous.

@JohnJunior there's a link at the bottom of the page, and I believe if someone found their way into the chatroom they'll know what math.se refers to

Then if $A = [0,\infty]$ and $f(x) = x$ and $g(x) = sin(x)$ then $fg$ is not UC.

.

@JayeshBadwaik Why is it not UC?

@anon Where at the bottom of the page?

The MATHEMATICS logo.

(It's written in caps, I'm not yelling.)

Thanks I just found it :)

@PeterTamaroff I guess one can prove it, but it will be long, so as an idea, Derivative is $x\cos(x) + sin(x)$ which is unbounded. So $f(x) - f(x_{0}) < \epsilon$ will depend on where $x_{0}$ is

while I work on the proof. I must have it somewhere in my notes though, wait let me search through it.

logo MATHEMATICS

@anon Thanks for the change in title :-D

@PeterTamaroff Okay, so it is here in my notes, the counterxample is from surprise, surprise Counterexamples in analysis.

@JayeshBadwaik Hm, OK. So the proof is pretty easy in the end.

@PeterTamaroff Get that book, whenever I feel that the condition I am using is not weak enough, I refer to it for counterexamples, it is a good resource that way.

@JayeshBadwaik You mean for any $\epsilon >\pi/2$ at first right?

@PeterTamaroff Yup, choose $\epsilon > \pi / 2$ and $x_{0} = 0$, then you get $\delta > \pi/2$. Once that is done, you can show that, for large $x_{0}$ $\delta > \pi /2 \nRightarrow \epsilon < \pi/2$ and then, you can show that for any $\delta$ then, at some stage $\epsilon > \pi/2$

@JayeshBadwaik Why do I "get" $\delta >\pi /2$?

$f(\pi/2)= \pi /2 * \sin(\pi/2) = 1$ after which the function starts decreasing.

Hmm, that's easy right?

@JonasTeuwen It is easy.

Shouldn't I choose $\delta$?

It is something about the growth, so you should take two almost not-uniform continuous functions.

I think I can see how to prove it.

Take any zero of the function and any peak point of it.

@PeterTamaroff yeah.

Then we can make that difference as large as we want, although the points are close.

Yup.

@JayeshBadwaik Is that what you're trying to tell me?

Yup.

@JayeshBadwaik =)

@PeterTamaroff =)

=)

:)

@JohnJunior Your MSE logo gravatar confuses me.

@JayeshBadwaik :0

@JohnJunior :P :P

:-D

hi @robjohn

:-D

@JohnJunior In a sense it is apt (the logo), users will feel MSE is welcoming them whenever they come to chat. Though you never welcome me. :-(

@JayeshBadwaik You are a trusted regular member of this room who needs no introduction.

@JohnJunior Ohh. :-)

@JayeshBadwaik I have a doubt. For the choice of $x,y$ we get that $|x-y|=\pi/2$ for every $n$.

And we get $|f(x)-f(y)|=\pi/2+\pi n$ for every $n$.

Yes. My point is as follows. If the function is UC then for $\epsilon > \pi /2$ you must have $\delta >\pi/2$ due to the example at $x_{0} =0$ and then you show that for every $\delta \geq \pi/2$, you can find $x_{0} , x_{1}$ such that $f(x_{0}) - f(x_{1}) > \pi /2$

@WillHunting But.. it is a prerogative for the test to be applied...

Now, I want to find some $\epsilon >0$ such that for some $x,y$, there is no $\delta >0$ making $|f(x)-f(y)|<\epsilon$

@JayeshBadwaik I don't understand the third sentence. What is the example at $x_0=0$?

@PeterTamaroff Okay, I will describe again. Just a sec. This time, without leaving any detail.

Yup, I am sorry about that, I just realized I was writing it incorrectly.

@JayeshBadwaik Do you ahve the page where this is examined in that book?

@unNaturhal Have you read this article on the infinitesimal?

I actually started out with that formulation to give a visual picture about why the function is not UC. But to prove it, you require different method. It is on page 26. (I thought I had linked to the page?)

@JohnJunior You've swiped our site icon... :-)

@JayeshBadwaik Bleh, but they don't prove it

@JohnJunior I was searching fot it! Thanks I will read soon!

@robjohn That's what the Raiders do best ;-)

@PeterTamaroff Yup they don't. I have proved it, but it is rather brief. I am reconstructing the details. What I actually did is I have a theorem where I have shown that if you can show that the derivative is unbounded, then you can find $\frac{f(x_{1}) - f(x_{0})}{x_{1} -x_{0}} = \frac{\epsilon}{\delta}> N$ for any $N$ and hence, for any $\delta$ I can find an $x_{1} - x_{0} < \delta$ such that the $\epsilon > M$, something like this, and I use that theorem everywhere.

@robjohn BTW were you in LA when we won our Super Bowl as the LA Raiders?

@JohnJunior So, when Leibniz, in his test, says: "Given $a_n \geq 0$ a sequence decreasing and infinitesimal. Then the serie $$(-1)^{n-1}a_n$$ converges" He means with infinitesimal that the sequence $a_n$ approaches $0$ as $n$ approaches infinity?

Okey :P