Mathematics

user19161

12:00 AM

@Link What do you mean by velocity cut into half?

Link

actually

intial velcoity

one sec

i'll show you a graph

user19161

@link Have you been kidnapped?

user19161

LOL

Link

no

k

here is the question:

How would graph A change if its accelration was kept the same but its intial velcoity was cut in half?

the line is graph A

user19161

@Link Acceleration is the rate of change of velocity with respect to time.

user19161

12:09 AM

So it is the gradient of the graph.

Link

okay

I know that

gradient?

user19161

So start from 3 instead of 6.

Link

oh

user19161

And the slope should be the same.

Link

thats it?

user19161

12:10 AM

Slope=gradient.

Link

wow

thanks

user19161

Wait.

user19161

As you can see, even the acceleration in the first graph changes.

user19161

So what does the question really mean?

Link

huh?

user19161

12:12 AM

Acceleration changes from negative value to zero (flattening out of the graph).

Link

yes

user19161

So I ain't sure what the question is.

Link

it's what I said

how would the line change

thats it

user19161

No, I mean it is not clear whether they want both graphs to have the same acceleration at all points.

user19161

I guess if you ask this question, you don't really know what I am asking here.

Link

12:15 AM

yea

wait

thats what it wants

user19161

Anyway I have said everything I can say here.

Link

to have the same accelration

but thanks

its a simple high school class

so thats probably it

user19161

Often school teachers don't know what they are talking about

Link

calc is a lot easier -.-"

user19161

@Link Calculus?

Link

12:17 AM

yep

ayways

if the slope of velcoity vs time graph is the same throughout

then the slope of an accelration vs time graph is just = 0 right?

Peter Tamaroff

@anon

user19161

@Link If slope of VT graph is the same, that means acceleration is constant, which means that slope of AT graph is zero, yes.

user19161

From displacement to velocity to acceleration, one just takes derivatives.

user19161

In the other direction, one takes integrals.

user19161

Derivatives correspond to slopes and integrals to areas.

Link

12:22 AM

okay

that just helped me so much

love the calculus way of descbing it

Peter Tamaroff

@MarianoSuárez-Alvarez Hola!

N3buchadnezzar

^^

Link

well bye guys and gals, and thanks for the help @WillHunting

anon

huh?

@PeterTamaroff what's up

Peter Tamaroff

@anon I'm stuck

anon

12:33 AM

I assume commas represent decimal points

Peter Tamaroff

@anon You assume well.

N3buchadnezzar

@anon Any quick pointers why $\sin x / x$ is analytic?

anon

@PeterTamaroff If $f(a+h)\ge10^{-n}$ and $h<10^{-n}$ then $f(a+h)/h>1\not\to0$.

@N3buchadnezzar Using only the side-over-hypotenuse definition?

You clearly need only check $x=0$, and it should be equivalent to showing the limit at 0 exists. Which is to say, the derivative of sin() exists at 0.

N3buchadnezzar

@anon No, by any means neccecary. And I know the problem is at the origo.

anon

But I assume you don't automatically have the Taylor expansion or the formula as a linear combination of complex exponentials. So essentially you start with only the triangle definition.

N3buchadnezzar

12:40 AM

The problem is indeed a bit backwards. The goal is to prove that $\lim_{x \to 0} \sin x / x = 1$ by using the taylor series. And in order to justify switching the limit and the sum, we need the function to be analytic.

Jayesh Badwaik

12:59 AM

@anon About that semigroups we were talking about.

user19161

@jayesh I saw your removed message. It implies continuity at the point but not on an interval containing the point.

user19161

I am going to sleep now. Have fun boys and girls!

Jayesh Badwaik

@WillHunting You are correct, I just found a counter-example to it.

math.tamu.edu/~tvogel/gallery/node4.html

Similarly, the ruler function may be differentiable at 0.

leo

hello :-)

Jayesh Badwaik

@leo hi.

leo

1:10 AM

@PeterTamaroff Imk

Peter Tamaroff

@leo I think I have proven $2.$ one way.

leo

@PeterTamaroff I guess "$\implies$"

Peter Tamaroff

@leo Well. Suppose that $\lim_{x\to a}f(x)=L$.

leo

yep

Peter Tamaroff

Then for every $\epsilon >0$, there is a $\delta >0$, such that, for all $x$, $0<|x-a|<\delta\implies |f(x)-L|<\epsilon$

leo

1:13 AM

yep

Peter Tamaroff

Let $\{x_n\}$ be any sequence with $x_n\to a$.

Then for every $\delta>0$ there exists an $N$ for which $n\geq N$ implies $|x_n-a|<\delta$

But for this $N$, we'll have then $$|f(x_n)-L|<\epsilon$$, from where the sequence $f_n=f(x_n)$ converges to $L$.

Am I left or right?

leo

@PeterTamaroff not only for that $N$ but for all $n\geq N$

and the everything is right

Peter Tamaroff

@leo Yes, that is what I meant.

leo

the "$\impliedby$" is true because it can't be false

@PeterTamaroff ok fine

Peter Tamaroff

@leo Proof by contradiction?

leo

1:17 AM

@PeterTamaroff indeed :-)

proof by tradition

:6177004 Tomae's function?

Jayesh Badwaik

@leo Yes, I was thinking. but then realized, f(0) = 1 right?

Peter Tamaroff

@leo So I should say "supongamos ahora que para cada secuencia $\{x-n\}$ con $x_n\to a$, la secuencia $\{f_n\}$ con $f_n=f(x_n)$, converge a $L$, pero que $\lim\limits_{x\to a} f(x)\neq L$"?

leo

@JayeshBadwaik I think is not even well defined at $0$, because $0$ is rational but $0=0/q$ with $q$ any natural number

@PeterTamaroff exacto!

Jayesh Badwaik

@leo Okay, hmm. Yup, hence my confusion. If we define it as $0$ at $x=0$, then it is differentiable.

Peter Tamaroff

@leo La idea es mostrar que la suposicion $\lim f\neq L$ es absurda con lo que debe ser $\lim f=L$?

Es casi trivial, no?

leo

1:29 AM

@JayeshBadwaik no. No matter how you define $f$ at $0$ you get a discontinuity at that point, so it can't be differentiable at $0$. She resist :-)

Peter Tamaroff

@leo O sea, creo que seria mas prolijo probar $A\implies B$ y $\neg A\implies \neg B$, no?

leo

@PeterTamaroff correcto

@PeterTamaroff así es

@PeterTamaroff pero me gusta más la prueba por contradicción en este caso

Peter Tamaroff

@leo OK.

revolviendo el tacho de basura

desdoblando bollito de papel

leo

@PeterTamaroff good

anon

@JayeshBadwaik eh?

leo

1:40 AM

@PeterTamaroff lo digo porque en esa forma todo empieza: "suponga que para toda sucesión $x_n\to a$ la sucesión $f_n$ asociada converge a $L$. Suponga que $\lim_{x\to a}f(x)$ no existe o no es $L$"

Peter Tamaroff

@leo Creo que la contradicción es tan obvia que no me doy cuenta.

leo

si $\lim_{x\to a}f(x)=K\neq L$ por la dirección que ya probaste se sigue que todas $f_n$ convergen a $K$ y eso es contradictorio

Jayesh Badwaik

@anon You remember I asked you about when can semigroups be isomorphic and stuff. Got this which was related.Hence, shared that with you.

Peter Tamaroff

@leo Asi de directo? XD

leo

@PeterTamaroff si $\lim_{x\to a}f(x)$ no existe entonces se puede cocinar una sucesión $a_n$ tal que la $f_n$ asociada no converja a $L$

@PeterTamaroff si. Es solo usar la dirección que ya tienes

Peter Tamaroff

1:45 AM

@leo La contradiccion seria en realidad que las secuencias convergen a dos limites distintos, que es absurdo, no?

leo

@PeterTamaroff eso estaba a punto de decir

piensa la otra. Voy a comer :D, ya vuelvo

leo

2:01 AM

@PeterTamaroff ya volví

Peter Tamaroff

@leo No soy muy buen cocinero aparentemente

leo

@PeterTamaroff Supongamos que $\lim_{x\to a}f(x)$ no existe. Entonces para cada número real $K$ existe un $\epsilon\gt 0$ tal que para cada $\delta\gt 0$, hay un $t$ con $|t-a|\lt\delta$ pero $|f(t)-K|\gt\epsilon$

Peter Tamaroff

@leo Si.

@leo Pero esa $l$ debe ser una $K$! =)

leo

ok

la cosa es para cada número real $K$. $L$ es un número real entonces apliquemos eso a $L$

Peter Tamaroff

@leo OK.

leo

2:10 AM

entonces existe un $\eta\gt 0$ tal que para todo $\delta\gt 0$ bla, bla,...

la cosa es para cada $\delta\gt 0$, en particular $1/n\gt 0$ para cada $n\in \Bbb N$

Entonces para cada $n\in\Bbb N$ existe un $t_n$ tal que $|t_n-a|\lt\frac{1}{n}$ pero $|f(t_n)-L|\gt\eta$

Peter Tamaroff

@leo Perfecto. Cuando vi el $1/n$ ya me di cuenta... =D

leo

ok

esa $(t_n)$ es la queríamos cocinar

Peter Tamaroff

@leo Pero entonces $\{t_n\}$ es una secuencia con $t_n\to a$ pero $f_n\not \to L$, que es absurdo.

leo

@PeterTamaroff sí, contradicción. Por lo tanto debe ser que $\lim_{x\to a}f(x)$ existe y es $L$

Peter Tamaroff

@leo B-E-autiful!

leo

2:17 AM

recuerda como empezamos:

37 mins ago, by leo

@PeterTamaroff lo digo porque en esa forma todo empieza: "suponga que para toda sucesión $x_n\to a$ la sucesión $f_n$ asociada converge a $L$. Suponga que $\lim_{x\to a}f(x)$ no existe o no es $L$"

Mohamed Ahmed Nabil

why is i^2 = -1?

Peter Tamaroff

@MohamedAhmedNabil What do you mean by "is"?

leo

@MohamedAhmedNabil why is $1\neq 0$?

Mohamed Ahmed Nabil

@PeterTamaroff can anyone give me the proof to

i^2=-1 ?

Peter Tamaroff

@MohamedAhmedNabil That is not a theorem. And if it is, then the definition of $i$ is $i=(0,1)$, and you have defined complex numbers as ordered pairs, and multiplication of complex numbers accordingly.

leo

2:20 AM

@MohamedAhmedNabil There is no proof. It is an axiom.

Mohamed Ahmed Nabil

@PeterTamaroff why are you writeing $ everywhere?

Peter Tamaroff

@MohamedAhmedNabil I guess you don't have CHATJax

@MohamedAhmedNabil Here

Mohamed Ahmed Nabil

@PeterTamaroff ty

leo

@MohamedAhmedNabil Or you can start with more primitive axioms and deduce it from it as Peter suggest

@PeterTamaroff hay mucho de eso que acabamos de hacer en análisis. De negar una hipótesis del tipo "para cada $\delta\gt 0$" se consiguen cosas para cada $n\in \Bbb N$ y así.

Son bonitas construcciones

Peter Tamaroff

@leo Si. Justo ayer me daba cuenta lo detallado que hay que ser a negar la existencia de un limite, al negar la continuidad o mas complicado aún, la continuidad uniforme!!!

@leo Si =)

@leo Es mas. Estoy intentando probar que $x\sin x$ no es uniformemente continua en $[0,\infty)$.

leo

2:26 AM

@PeterTamaroff esa o $x\sin (1/x)$ ?

Peter Tamaroff

@leo $x\sin x$.

Porque (supongo) probe que si $f$ y $g$ son unif continuas y acotadas en $A$ entonces $f\cdot g$ es u.c. en $A$. Quiero probar ahora que la hipotesis de que ambas sean acotadas es necesaria.

La idea es que si $x=n\pi+\pi/2$ e $y=n\pi$ entonces $|x\sin x-y\sin y|=n\pi+\pi/2$

Y $|x-y|=\pi/2$

leo

correcto

Peter Tamaroff

Es decir, tomamos una raiz y un valor pico.

Entonces, la negacino de u.c. sería
"Para *cierto* $\epsilon >0$, se tiene que *para todo* $\delta >0$ existe *algun* $x$ y *algun* $y$ para el cual $|x-y|<\delta$ pero $|f(x)-f(y)|>\epsilon$??"

leo

@PeterTamaroff eso es correcto

Peter Tamaroff

@leo OK. Ahora tengo que aplicarla.

Mi problema es que lo mio funciona para todo $\delta >\pi/2$!

leo

2:36 AM

@PeterTamaroff si ese es el problema.

Peter Tamaroff

@leo Lo tendré que ver mañana, ya es tarde...

leo

@PeterTamaroff pero pa probar que la acotación de ambas es necesaria, $x\sin(1/x)$ te sirve, puedes hacer algo parecido. Es que esa tuya no estoy seguro.

Transcript for

$Mathematics$ Mathematics