The h Bar

12:00 AM

yet we only integrate once to get $V$ in the second equation

why is this?

I think of the laplacian as taking the derivative twice. To "undo" this, it only seems intuitive to have to integrate twice...

heather

12:19 AM

question about scanning tunneling microscopes - anyone around?

let's say i wish to move a molecule of adenine from a "cartridge" to a specific location elsewhere away from the cartridge, can a scanning tunneling microscope do that? because all the info i've read about the stm seems to imply it's moving the atoms/molecules within the substance.

0celou7

@heather are you good at circuits?

heather

@0celou7, depends on what you define as good. probably not

i know about components of circuits, and i can solder, and do some basic calculations.

0celou7

aha, these are amps, not miliamps

never mind

I'm getting 3.18F for a capacitor but that's ok given the massive amperage

maybe

DanielSank

1:48 AM

Why do so many people write "My question is..."?

Just ask the damned question.

> My question is, why is the sky blue.

Versus

> Why is the sky blue?

Down with "My question is"!

@0celou7 "Amperage" is a word used by electrical engineering hillbillies. The word you're looking for is "current".

0celou7

@DanielSank I am an engineer you know

DanielSank

Yes, but you don't need to sound like an uneducated buffoon.

"Amperage" is like calling distance "meterage".

0celou7

@Slereah Sharpe, in his book on Cartan/Klein/Projective geometry, references Napier's 1614 table of logarithms...

DanielSank

The only acceptable case of doing that is "voltage".

0celou7

@DanielSank I'd call it footage.

DanielSank

1:50 AM

@0celou7 Heehee

clever

0celou7

@DanielSank I don't know, calling it amperage seems natural...

we also say wattage

British thermal unitage

etc.

DanielSank

@0celou7 Nope.

Power, not wattage.

0celou7

> watt·age
ˈwädij/
noun
noun: wattage; plural noun: wattages
a measure of electrical power expressed in watts.
the operating power of a lamp or other electrical appliance expressed in watts.

sorry Dan, looks like I win

0celou7

2:26 AM

@DanielSank you know that $\cos t^2$ integral is pretty nice

Physics Meta

0

Q: Interpretations of questions

How do we interpret a question correctly when there exits multiple intepretations? for example in this question here i interpreted it one way and answered, but apparently some others(not the OP) downvoted it saying it dosen't have enough information. Can someone suggest ways to deal with such mi...

discussion

0celou7

It gives a nice example of an $L^2(\Bbb R)$ function that does not go to zero at infinity

(if you take the square root first)

satyatech

Guys here is a question on rotational dynamics .Please understand this easy question and tell me why the time of flight is used here.

0celou7

is this for JEE?

Fawad

@0celou7 no,for increasing problem solving skills ;)

satyatech

2:41 AM

Fawad ,where is diameter and how will it fly?

Fawad

@satyatech let's discuss here

0celou7

Donald Trump: "Get 'em outta here!"

►

Fawad

@satyatech I don't know,how about asking on physics.SE or waiting for John Rennie to come.

0celou7

Why does everyone want to wait for @JohnRennie ?

Fawad

@0celou7 there is high probability that he will answer correctly

satyatech

2:51 AM

@Fawad , I understood the problem,thanks for giving time..

Fawad

@satyatech why it took time of flight equation?

TrY iS CheM

3:05 AM

hi @Fawad

any idea how to do it

Slereah

@0celou7 Noice

0celou7

3:23 AM

@Slereah he credits Napier with discovering Lie groups since $\exp:(R,+)\to (R_{> 0},\times)$ is a Lie group isomorphism

satyatech

So u are prep for her @TrYiSCheM

*JEE = her

0celou7

out out out!

Slereah

@0celou7 isn't addition a Lie group with negative numbers

Or $\Bbb R^+$ with division

0celou7

I have all of R on the left

Slereah

The ancient greeks only worked with $\Bbb R^+$

surely they should get the credit

0celou7

3:30 AM

did they really have a notion of all of R+ tho

Slereah

Though I think Rovelli still wins for having Plato in his bibliography

0celou7

did they understand it was complete

Slereah

I don't think they had a notion of "complete"

I think at best they had a notion of algebraic numbers

0celou7

@Slereah If I write a book I'm putting cave paintings on the cover and referencing a neanderthal

Slereah

Since they had the notion that $\sqrt[3]{2}$ wasn't constructible

Who was the first to discover a non-algebraic number and knew it was one, anyway

"The name "transcendental" comes from the root trans meaning across and length of numbers and from Leibniz's 1682 paper in which he proved that sin(x) is not an algebraic function of x.[1][2] Euler was probably the first person to define transcendental numbers in the modern sense.[3]"

"Johann Heinrich Lambert conjectured that e and π were both transcendental numbers in his 1768 paper proving the number π is irrational, and proposed a tentative sketch of a proof of π's transcendence."

"Joseph Liouville first proved the existence of transcendental numbers in 1844,[5] and in 1851 gave the first decimal examples such as the Liouville constant"

noice

Yashas Samaga

3:34 AM

how do I retract a comment flag? :/

0celou7

you don't

Yashas Samaga

aw

0celou7

you live with your poor decision

Yashas Samaga

aw x2

Slereah

although the ancient greeks did come up with the notion of infinite sequences

TrY iS CheM

3:43 AM

@satyatech preparing

Yashas Samaga

which one?

TrY iS CheM

@YashasSamaga

Yashas Samaga

20?

The Ka is given for RNH3+

TrY iS CheM

sorry it is 0.2

Yashas Samaga

its 0.2

oops

Therefore, the reaction is: RNH3+ -> RNH2 + H+

Ka = [RNH2][H+]/[RNH3+]

pH = 4 => [H+] = 10^-4

Ka = 2x10^-5

2x10^-5 = 10^-4 * ([RNH2]/[RNH3+])

10^-4 cancels both sides

you get 2x10^-1 on the left

the right side has your ratio

2x10^-1 is 0.2 :P not 20

This is how I lose marks :D

TrY iS CheM

3:58 AM

oh great

thanks

@YashasSamaga this happens with everyone

0celou7

not with Feynman

Yashas Samaga

:O @0celou7 is Feynman fan?

0celou7

no

Yashas Samaga

aw

0celou7

but he didn't make trivial mistakes

Yashas Samaga

4:02 AM

Feb 16 at 10:33, by Yashas Samaga

Feynman's lectures on physics is a good enough novel for me lol

Yashas Samaga

4:13 AM

I wrote a program to calculate the value of e using probability and derangements

0celou7

4:40 AM

@YashasSamaga do you want to try a calculus problem?

Yashas Samaga

I'll decide after I see the problem :P

0celou7

@YashasSamaga Consider the function $f$ which is continuous on $[0,1]$ and never negative. Suppose $\int_0^1f(x)\, dx=0$. Is $f(x)=0$ everywhere?

Slereah

I think so?

Yashas Samaga

yea

Slereah

Isn't there a theorem about how for those cases, the integral is between the two boundaries

0celou7

4:43 AM

@YashasSamaga can you justify your answer?

Yashas Samaga

If f(x) goes a little up for an interval dx

then you are going to have a finite area

Slereah

mean value theorem

or something

Yashas Samaga

as f(x) can never go negative

you can never cancel it out

0celou7

that's the right idea

I wanted to see if JEE teaches you rigorous proofs

Slereah

Ah yes

$$\min f(x) \leq \int_a^b f(x) dx \leq \max f(x)$$

Yashas Samaga

4:45 AM

tbh I don't forgot Mean Value theorem and Roll's theorem

I don't remember the definition

when it was taught, it sounded obvious

0celou7

@YashasSamaga they are actually very hard to prove

Slereah

Hm, does that help actually

not really

0celou7

it's hard to convince someone that they are hard to prove because they seem so obvious :D

Yashas Samaga

I got a rigorous proof for your Q

Slereah

I mean it's pretty obvious from the Riemann sum

Yashas Samaga

4:46 AM

I can write the integral as a summation of infinite pieces

0celou7

@YashasSamaga no

Slereah

Or the Lebesgue sum, I guess

0celou7

that's not rigorous

Yashas Samaga

huh

Slereah

It's totally rigorous

if u use the Riemann integral

0celou7

4:47 AM

I guarantee anything coming after "write the integral as a summation of infinite pieces" will not be rigorous.

Yashas Samaga

ye

nope

ye was @Slereah and the nope was at @0celou7

You can write the integral as : f(0)d + f(h)d + f(2h)d.....

where d = (upper limit - lower limit)/n

Slereah

well not quite

Yashas Samaga

in this case d = 1/n

Slereah

I mean you have to at least write it rigorously :p

Yashas Samaga

leme write & upload a pic

0celou7

4:50 AM

@Slereah The quick proof is to recall that $f\ge 0$ and $\int f=0$ implies $f=0$ a.e., which for continuous functions implies $f\equiv 0$ since a.e. statements hold everywhere for continuous ones.

There is an easy direct proof for Riemann integrals but the details are tricky.

Slereah

Conway base 13 function

The Conway base 13 function is a function created by British mathematician John H. Conway as a counterexample to the converse of the intermediate value theorem. In other words, even though Conway's function f is not continuous, if f(a) < f(b) and an arbitrary value x is chosen such that f(a) < x < f(b), a point c lying between a and b can always be found such that f(c) = x. In fact, this function is even stronger than this: it takes on every real value in each interval on the real line. == The Conway base 13 function == === Purpose === The Conway base 13 function was created as part of a "produce...

that's a weird ass function

0celou7

lol

Yashas Samaga

*a-b shud be b-a

I forgot to add few assumptions

0celou7

I think you mean "nonnegative" where you had "positive"

Slereah

What is a parallely propagated tetrad

Or more to the point

what's a non-parallely propagated tetrad

0celou7

4:56 AM

A parallel frame

Yashas Samaga

yea I made too many typos

Slereah

is that just a condition on the coordinates

Yashas Samaga

There are over a dozen mistakes.

I mixed the general case with your specalized case

Slereah

a baker's dozen

0celou7

@Slereah Ahh, one uses the topological definition of continuity...

quite tricky

No, that still doesn't work

Yashas Samaga

5:00 AM

"where h is" some useless shit

ignore it

0celou7

@YashasSamaga So you're assuming that $f(x)>0$ somewhere in $[0,1]$, right?

Yashas Samaga

?

0celou7

What are you trying to do here

Yashas Samaga

"As each term in the sum is positive" must be "As each term in the sum is non-negative"

Tim The Enchanter

The integral will be nonzero if the function f is continuous over any open interval (a,b). So for the integral to vanish, f can only be nonzero at isolated points, the fact that f is continuous leads to a contradiction.

0celou7

5:03 AM

@TimTheEnchanter hush

Tim The Enchanter

silenced

0celou7

@YashasSamaga Your proof doesn't really work, because that sum is only zero in the limit. And in that case it doesn't make sense to speak about $f(1/n)$ or $f(2/n)$ or whatever being zero

that's the problem with Riemann sums (and why we don't use them in pure math), you're taking an infinite number of terms but the terms themselves depend on the number of terms!

it's quite annoying

Slereah

Aren't Lebesgue integrals also basically sums, anyway

since they're only properly defined for integrating constant functions

0celou7

In any case, here's the proof: Assume $f(x)>0$ for some $x\in[0,1]$. Then by continuity there is $\epsilon>0$ such that for $|x-y|<\epsilon$, $f(y)>\frac{1}{2}f(x)$.

So take a little rectangle on the set $|x-y|<\frac{1}{2}\epsilon$ with height $\frac{1}{2}f(x)$.

user228700

@JohnR: Morning :-) Do ping me when u're free!

0celou7

5:09 AM

The area of this is $\frac{1}{2}f(x)\epsilon$, but it lies underneath the graph of $f$.

Hence $\int_0^1f(x)\,dx\ge \frac{1}{2}f(x)\epsilon>0$, which is a contradiction.

Maybe replace that $x$ with $x_0$ for clarity.

@Slereah you can define them by limits of sums, sure

but you use more general measurable sets instead of rectangles in the definition

John Rennie

@Kaumudi.H ping ...

user228700

@JohnRennie Hi :-) I have a quick question about optics and sign convention...

Slereah

@0celou7 does it make much of a difference from a practicle standpoint for all but the most pathological functions

John Rennie

@Kaumudi.H Ok, not my strongest area but ask anyway.

user228700

@JohnRennie OK. I have always been extremely confused about this. After having derived a given formula using sign convention, we have to apply it again while substituting values. How come?

0celou7

5:17 AM

@Slereah the big difference is for convergence, not really the types of functions you can integrate

And that's very important, it's what makes L^2 a Hilbert space

Slereah

How do you show that say $\int_a^b x dx$ converges

John Rennie

Well if you derived your equation by assuming (for example) that light moving to the right is positive then the numbers you plug into the equation have to follow the same convention. Or have I missed what you're asking?

Yashas Samaga

@Kaumudi.H physics.stackexchange.com/questions/240743/…

0celou7

@Slereah I mean convergence of integrals of a sequence of functions

I have to go to bed, later

user228700

@YashasSamaga Lol, I had that same question before but I don't care too much about the derivation now, only how to apply it in problems. ::Reading ur answer::

user228700

5:21 AM

Yep, that makes sense. Thanks!

Yashas Samaga

5:34 AM

I was just wondering if the refractive index for a mirror could be -1.

It worked for 4 formulae.

Tim The Enchanter

Negative-index metamaterial

Negative-index metamaterial or negative-index material (NIM) is a metamaterial whose refractive index for an electromagnetic wave has a negative value over some frequency range. NIMs are constructed of periodic basic parts called unit cells, which are usually significantly smaller than the wavelength of the externally applied electromagnetic radiation. The unit cells of the first experimentally investigated NIMs were constructed from circuit board material, or in other words, wires and dielectrics. And in general, these artificially constructed cells are stacked or planar and configured in ...

yuggib

7:14 AM

@0celou7 the l.i.m. notation is used to intend the limit in the $L^2$-norm. And of course the limit in the $L^2$ norm can be reformulated as an almost everywhere convergence, since $L^2$-functions are equivalence classes of almost everywhere equal functions.

skill patrol

\o @yuggib

yuggib

7:42 AM

@skillpatrol o/

skill patrol

7:59 AM

Have you watched any of the March madness so far? @yuggib

user228700

8:33 AM

@Ken: Hi, boss :-P How's it going?

Kenshin

Sup girl

things are going well

how bout u

user228700

I see. Eh, my exam's in a week. When u going to...Japan, was it?

Kenshin

yeah

soon

user228700

Cool :-) Extended trip or just a quick vacation?

Kenshin

yeah just a vactation

vacation

r u ready for ur exam Kaumu?

user228700

8:36 AM

Nice. U haven't been online as much. Been busy, famous and rich person? :-P

yuggib

@skillpatrol nope

user228700

@Kenshin ...don't ask me.

Kenshin

yeah busy with lots of stuff

I think u will do fine Kaumuj

I believ in you

user228700

Cool. Thanks a lot, man :-) Have fun on that trip! How's the website?

Kenshin

havent really been monitoring it

but I htink people are still asking and answering questions there

user228700

8:39 AM

...has the crowd grown?

Kenshin

I dunno not by too much

It needs better SEO

@Kaumudi.H u should see beauty and the beast after your exams

great movie

user228700

Ohh, OK...

user228700

Oh, really? Will do! :-)

user228700

Have u watched any other movies?

Kenshin

yeah

I also watched Logan

user228700

8:40 AM

Riight. How was that?

Kenshin

yeah not bad

but beauty and the beast is better

user228700

Wokay. Not much auto-tune, I hope?

Kenshin

lol

na really good cast they can singw ell

hermonie played belle really well

user228700

Phew. OK. Some people were complaining about the auto-tune.

Kenshin

h relaly

well i ddin't notice if there was any

ppl can complaingn about anythign lol

user228700

8:42 AM

...I'm not into Disney movies but I'll certainly watch this one because I may be in love with Emma Watson :-P (Hyperbole alert)

user228700

@Kenshin Lol, true.

Kenshin

ooo la la

@Kaumudi.H so did your poem get played

user228700

No, not yet :-/ But there is hope...

user228700

Lunchtime, man. I'll ttyl. Have fun on that trip! Take care :-)

Kenshin

laterz ty

goodluck for ur exams if i don't talk b5

b4

DanielSank

9:55 AM

@0celou7 math.stackexchange.com/questions/2205058/…

Bernardo Meurer

10:40 AM

@KyleKanos ::pokes::

Balarka Sen

yo

when does 0celo7 have an extra u in his name

John Rennie

2 days ago, by 0celou7

@OBE I embraced the French in me

Balarka Sen

ahaha, I see

I prefer Russian. 0celovsky7

2

Bernardo Meurer

10:58 AM

I like him better as 0PROPRIETARYMALWAREcelo7

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