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12:09 AM
hello
 
Howdy
 
 
1 hour later…
1:10 AM
@SirCumference I'd suggest considering whether complying with those terms would actually pose a problem for you, before deciding you can't use them.
SE's designs are all done in-house, so they own the copyrights
 
rob
1:30 AM
People of earth, I greet you
 
people and AIs of outer space, I greet you
 
may the force be with you
 
What about AIs of earth? :(
 
AIs of earth, I'm sorry you were left out. we greet you as well. please don't kill us all. =)
 
2:04 AM
@DavidZ while I understand why you desensationalized the cheese question, :-(
@heather wat?
 
@DanielSank it was meant to be a response to rob's comment.
 
 
4 hours later…
5:48 AM
Hello.
 
user228700
@Swapnil: Hi :-) Happy Republic Day!
 
Happy Republic Day to you too :)
India will actually become independent when it gets rid of JEE, though :P
 
user228700
^ That's not happening anytime soon :-P
 
Lol. You're appearing this year, no?
 
user228700
Yeah.
 
5:55 AM
Best of luck, kill it :)
 
user228700
Thank you :-)
 
@JoshuaLin My regards to an IPhO contestant :)
 
user228700
@JohnR: Hihi! :-)
 
Morning :-)
Another adventurous morning, or just a regular one? :-)
 
user228700
Oh, just a regular one. Watched a bit of the Republic Day parade on my T.V.
 
6:05 AM
I miss those :(
 
user228700
What dyou mean you miss them? U could've watched it on the T.V today. I think it's winding up now.
 
Study time. I was studying potential equilibrium :P
 
user228700
OK.
 
<miss them>: Miss those childhood days when I could watch them.
 
6:23 AM
Good morning
Hii @JohnRennie
Hope you are free now
 
Morning.
I did look at your question, but it isn't one I can help with.
 
Good morning.
 
Morning :-)
 
@JohnRennie why
 
I haven't done circuit theory for over 30 years, and can't remember any of it. Sorry.
 
6:26 AM
Post it on main.
 
@thoughtforfood do you saying it to me
 
yes
The main site is worth a try :-)
 
@thoughtforfood it would get closed as homework on the main site
 
Will it be answered as I don't know much about it
@JohnRennie if I post the solution also , then also will it be considered as homework
 
@koolman Yes, I'm afraid so.
 
6:34 AM
Okay :-\
@JohnRennie I have similiar type of question . May be you can answer it (not much about circuit)
May I ask
 
Yes, it's always worth asking. I'll answer if I can.
 
In this how they have written number of photons incident per second and i$_p$
@SwapnilDas Happy Republic day
 
@SwapnilDas Yo! Regards to an Ipho contestant indeed!
 
@koolman they calculate the total energy per second and get $258 \times 10^{-5}$W. Are you OK with that bit?
 
@JohnRennie yes
Wait a min
 
6:50 AM
OK. Then the number of electrons per second is the power divided by the energy per electron. They give you the energy in electron volts, $3.1$eV, and you get the energy in Joules by multiplying by the electron charge. So the energy per photon is $3.1e$.
 
@JohnRennie do they have written $\phi$ as the phase difference between two maxima is $\pi$ and distance between them is $\omega$ . x is the distance from origin
 
I get 3.1eV as the energy of the 400nm light ...
 
@JohnRennie okay , after that
@JohnRennie Sorry calculation mistake
 
The quantum efficiency of the detector is $20$% so the number of electrons produced per second is the number of photons times $0.2$.
 
Yes
 
6:57 AM
Then just multiply by the electron charge to get the current in coulombs per second i.e. amps.
 
Okay
 
Oh god...
I just woke up from a nap
 
@JohnRennie do you also have some idea how they have written the next circuit equation
 
I'm so screwed with everything I have to do...
 
And does my explanation of $\phi$ is correct @JohnRennie
9 mins ago, by koolman
@JohnRennie do they have written $\phi$ as the phase difference between two maxima is $\pi$ and distance between them is $\omega$ . x is the distance from origin
 
7:17 AM
The fringe width, i.e. the difference between maxima, is $\omega = 4$mm, and there is a phase difference of $\pi$ between each maximum.
 
Yes
 
So for any distance $x$ from the centre of the fringe the phase difference at the point $x$ is $$\frac{x}{\omega} \pi $$
 
Okay
Can phase difference be negative
 
Yes
 
I see
Do you get any idea of that equation
 
7:21 AM
Did you see this @koolman
 
@thoughtforfood are you also preparing for it
 
@koolman I'm not sure what you are asking there ...
 
@JohnRennie in the second question how they have written circuits equation . That is $(i_p -\frac{dq}{dt})R= \frac{q}{C}$
 
Don't know. I can't remember any circuit theory. Sorry.
 
No problem
 
7:26 AM
hey guys
how goeth it all
 
hello, fine thanks...you?
 
And in previous question , what is e in the equation of i$_0$
 
Hey guys, is there any difference between Plancks Law in terms of spectral radiance and spectral flux? I would assume there isnt
wow
the second i say something everybody just decides to disappear
 
How do you define "flux"?
I think your answer would depend on that :-)
 
7:48 AM
Last night dream:
Summary: I have a chemistry exam, however I overslept and thus I am late for it. Frustrated, I go back to sleep, and then when waking up again, found myself 1 hour earlier in the bed in the past. I then tried to use that (and with some luck, as the time travel is random) to study for all my exam topics. The dream however ends in a cliffhanger at the underground where the clock reads only 2 minutes before I will be late for the exam
The time travel in this dream is quite interesting, as it seemed to drawn its ideas from The Time Traveler's Wife and quantum mechanics

Basically, the way how this time travel works is you usual wavefunction in position space problem, except it is not position but time. Time also (somehow) became an observable and thus everytime you wake up, it's like this "temporal wavefunction" is being measured, and you randomly end up in one of the "event eigenstates". Meanwhile, going to sleep on a certain horse riding machine listening to a certain music brought you into a superposition with other ei
 
@koolman $i_0$? $e$ is normally the electron charge $1.602 \times 10^{-19}$ coulombs.
 
@Skyler have you seen this?
 
@JohnRennie oh yeah
@JohnRennie thanks a lot .
You are great
 
You're welcome :-)
 
8:05 AM
 
Dreamer, you know you are a dreamer
Well can you put your hands in your head, oh no!
:-)
 
Well, in fact, there are many real life examples if $t$ is not time, but some other observable. All you really need is an n level system (such as some ions) with the probability density skewed towards one direction of the system
Quantum dots (QD) are very small semiconductor particles, only several nanometres in size, so small that their optical and electronic properties differ from those of larger particles. They are a central theme in nanotechnology. Many types of quantum dot will emit light of specific frequencies if electricity or light is applied to them, and these frequencies can be precisely tuned by changing the dots' size, shape and material, giving rise to many applications. In the language of materials science, nanoscale semiconductor materials tightly confine either electrons or electron holes. Quantum dots...
I suspect a quantum dot with the electron density skewed towards one of the spatial direction will make a good analogy
 
So, hypothetically, and highly speculatively, what is shown in the dream might be an "anisotropic quantum dot in time"
 
have you heard the song?
 
8:15 AM
nope
1974, sounds very old to me
 
classic rock
 
was eating dinner
@thoughtforfood uh, i actually came here from that page
 
it gives the spectral <--> radial mathematical relation but not flux <-->radiance
i mean by unit analysis i can see whats going on
 
8:22 AM
radiance comes out to the observed flux at a specific cross-section
 
@thoughtforfood Supertramp? People still listen to Supertramp these days?
 
so as the cross-section is further out the radiance drops
 
Sure @JohnRennie quality never gets old.
 
but i dont really see any distance dependence in plancks equations
as listed in wikipedia (which is for the radiance)
Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T. The law is named after Max Planck, who proposed it in 1900. It is a pioneering result of modern physics and quantum theory. The spectral radiance of a body, Bν, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. Planck showed that the spectral radiance of a body at absolute...
 
@thoughtforfood Crime of the Century is a good album, as is Breakfast in America. Their other albums I can take or leave. I have those two albums on vinyl in a box in the attic - I bought them when they were first released :-)
 
8:25 AM
cool @JohnRennie
 
ferrimagnets do not conduct electricity? o_0
 
8:45 AM
If A and B are both part of the same universe, I don't see a reason why they cannot be connected in the most straightfoward way possible
e.g. consider two locations A and B that are 5 m apart, then before we plug a wormhole at A and B, its all just flat spacetime connecting A and B
Unless we are talking about that A and B are so far away that they are beyond each's hubble horizon, hence causally disconnected, then it might be possible they are connected by some nontrivial topology or even disconnected
So are we implying A and B are so far away that any information about their connectedness is not available?
 
9:02 AM
@Secret my point in that answer is that GR tells us nothing about the global topology i.e. how the flat spacetime region behaves between the two mouths of the wormhole.
 
Ah I see, and scifi just assume it is simply connected
 
9:28 AM
@JohnRennie where did you go in Cambridge?
 
I did my degree at Peterhouse then moved to Jesus to do my PhD
 
@JohnRennie Nice. I was at Magdalene 90-93. Thanks for reopening my question physics.stackexchange.com/questions/307541/…
 
You're welcome :-) I can see why it got closed because the other question is superficially similar.
I'm wondering if it's really true that cheese heats faster. It's a long time since I've heated anything with cheese ina microwave
 
@JohnRennie Me too although an eye for precision shows it is different. Is it possible to roll back Diracology's changes? More people would take an interest in physics if wordings like mine were permitted.
@JohnRennie And yes, put a piece of cheese on your dinner next time it goes in the microwave and when you take it out the cheese will be sizzling
@JohnRennie I don't think physics and fun should be mutually exclusive.
 
It's your question, and the rule is that the poster of the question has the final say. Just click where it says edited xx hours ago and roll back to the version you want.
 
9:33 AM
ok thanks
 
However there was some debate about the title amongst the moderators and I would be inclined to leave it as it is now.
I would guess the answer is indeed that cheese has a high conductivity and low specific heat, so it absorbs a lot of power and temp rise per joule absorbed is quite high.
But I don't know of any experimental data to back this up.
 
whoops already did it :(
@JohnRennie I'm sure there is more to it than. I think the key is that the absorption is greater, which will be down to molecular structure but this is the kind of insight you will only appreciate upon gathering experimental data.
 
@RobertFrost I suspect salt water (i.e. most sauces) absorbs microwaves more strongly than cheese but it will have a higher specific heat so the temp rise is slower.
The absorption isn't really down to the details of the molecular structure. Microwave absorption is effectively black body absorption and not tied to specific excitations of molecules.
 
@JohnRennie Your first assertion can be tested.
@JohnRennie Your 2nd assertion - does this marry up with the fact that molten glass absorbs microwaves more efficiently that cold glass?
 
See:
7
Q: Does a domestic microwave work by emitting an electromagnetic wave at the same frequency as a OH bond in water?

hawkeyeI was told once that microwaves work by exciting water molecules in food. Also that this worked because the frequency in the microwave was the same as that in the bond between Oxygen and Hydrogen in the water molecule. This similar frequency match lead to the molecule being energised, releasing h...

The absorption mechanism is dielectric heating
 
9:44 AM
@JohnRennie Molten glass absorbs microwaves more efficiently than solid glass
 
I can't comment on microwave absorption by glass, but it makes sense that the polarisibility of molten glass will be higher due to the increased mobility in the melt.
@RobertFrost you can roll back again to the revised title if you want. But then whether anyone really cares is debatable.
 
@JohnRennie Then maybe this is the answer. Perhaps this would suggest that materials whose molecules are mobile but offer greater resistance to motion, might heat up more than fluids?
 
Yo @JohnRennie, what's with the unilateral reopen? The cheese questions are identical and have identical answers.
Is there some chat consensus above about this?
@JohnRennie I completely disagree. The initial title is pure and clear clickbait and I think it is inappropriate for this site.
Also, @RobertFrost, for future reference, rolling back specific edits made by a mod is not particularly OK.
 
@EmilioPisanty shrug. You can edit again but the OP has the final say.
@EmilioPisanty No, the OP always has the final say. If the OP's version is unacceptable the question can be locked or deleted but otherwise it's down to the OP.
 
9:50 AM
@JohnRennie I do not want to start an edit war. But the least you could do before encouraging rollback of moderator edits is ping the corresponding mod, @DavidZ.
 
I didn't roll it back, the OP did.
 
@JohnRennie And indeed I voted to close, because it is a duplicate and does not add anything to the site that wasn't already present.
but funny if my closevote didn't mysteriously evaporate.
3
 
The linked question is about how a pizza cools down. How is that related to how something heats up in a microwave?
 
@JohnRennie the answers are pretty much identical. I only see minor wording differences.
You might disagree, but it is not the sort of clear-cut case that dupehammer is meant to address.
 
OK so the answers are duplicates, but the questions aren't duplicates.
 
9:52 AM
@EmilioPisanty "closevote evaporation" sounds so...Trump-like :D
 
At this point, frankly, I think the question does call for a rollback to the last moderator edit, closure and an edit lock.
Having said that, though, I have raised my flag and I will leave it for other people to decide.
But I do ask that you refrain from making unilateral moderation moves which you know are controversial.
(within which I'm including encouraging users to add clickbait to the site.)
 
Is there a meta question about "click bait"?
 
@EmilioPisanty My question physics.stackexchange.com/questions/307541/… is different to the prior one because it provides provision for the cheese to be more absorbent to microwaves per unit of heat capacity than other food, which the other does not.
@EmilioPisanty the other question covers the topic specifically of how hot the cheese feels irrespective of its actual temperature.
@EmilioPisanty in respect of the "clickbait" question. I had no intention to attract clicks. I think more kids would take an interest in physics if it were permitted to be fun and a fun obvious superlative "world's most powerful" transforms to a very dry subject into the awe we have all experienced when we get a cheesy lunch out of the microwave.
 
10:10 AM
Let the mods handle it @RobertFrost
 
@RobertFrost I don't think you understand the criticism, but I stand by what I said: your title turns a reasonable question into a junk-food one, and there is clear consensus (cf. ACuriousMind's comment above) that it needs to be changed.
@RobertFrost Regarding the duplicate, if your question is closed as one and you disagree, you need to edit your question to reference the earlier one and explain within the body of the question why it is different.
This procedure still holds, despite the fact that John has blatantly disregarded it.
That said, I am stepping aside for moderators to handle this.
 
@ACuriousMind In your answer [to the post](http://physics.stackexchange.com/questions/133121/negative-energy-levels-in-the-diagram-for-a-hydrogen-atom) you state "that a free, unbound electron has zero energy (that's convention you could just as well put another number there)."
Are you considering an unbound electron at rest? If not, then what about the variable kinetic energy it could have? Thanks.
 
@Alex yes, the reference point is the energy of an electron at rest at infinity
 
@JohnRennie Thanks, just wanted to confirm that.
 
10:34 AM
Hmm...interesting choice that 3700 K @JohnRennie no?
 
@thoughtforfood is there something special about 3700K? I assumed the OP just chose some random hyperbolically high temperature.
 
@JohnRennie Tungsten -- the metal with the highest melting point, melts at roughly 3700k at normal atmospheric pressure.
 
@JohnRennie The way you are stating that though sounds as if the only way to ionize a hydrogen atom is by taking it an infinite distance away from the nucleus?
 
@Alex Strictly speaking that's true, though in practice any macroscopic distance is essentially infinity. After all hydrogen atoms are pretty small :-)
 
I guess that's a little known physical chemistry fact :-)
 
10:42 AM
@thoughtforfood you'll have to ask the OP if that was his motivation ...
 
@JohnRennie So we are negotiating infinity with hydrogen atoms...Something doesn't sit well with that but if you say so then maybe it's fine.
 
This is curious: why isn't this terrible question in the reopen review queue?
It was edited after closure but I don't get a link to review it and it's not in the history.
 
@Alex the force an electron feels in a hydrogen atom is ke^2/r^2, where r is the distance from the nucleus. Ionising the atom means removing the electron to a distance where the force is zero, and that requires r to be infinity.
In practice the electron is ionised when its total energy, kinetic energy plus (negative) potential energy, is zero or greater and we generally don't worry whether it's actually at infinity or not.
 
@Alex No. What you need is for it to have enough energy for it to (a) reach infinity, and (b) have nonzero kinetic energy when it gets there, i.e. in the asymptotic limit.
You consider it to be ionized as soon as it has enough energy for that to happen.
 
@JohnRennie But from the pic it seems that if it is not out to infinity then it is technically in a bound state. Are you saying that as an experimenter, in practice, you can apply enough force on a particle in some energy level to increase it's kinetic energy so that it's total energy is greater than zero, hence we consider that ionized regardless of where it is?
So basically giving it enough force to reach an escape velocity?
 
10:54 AM
@Alex Yes. If the total energy is > 0 then the electron is unbound. If the total energy is < 0 it is bound. This applies to all bound states e.g. comets in the Solar System or protons in a nucleus.
You do get very loosely bound atomic states where the electron energy is very small but still < 0. These are called Rydberg states.
 
@JohnRennie Okay I understand thanks.
@JohnRennie Intuitively then it seems that the idea is that if it has $E > 0$, that even though there is still a force that decreases the kinetic energy, by the time the $E = 0$, it would be infinitely far away and hence unbounded?
 
The total energy E is the sum of the kinetic and the (negative) potential energy. To move the electron away requires increasing the potential energy (i.e. making it less negative) and this is done by decreasing the kinetic energy. So if E >= 0 that means when the PE has risen to zero the KE is still >= 0.
 
11:34 AM
@JohnRennie Do you have any idea how they actually would go about decreasing the the kinetic energy in practice?
 
@Alex I'm not sure what you mean. Are you asking how the kinetic energy of an electron decreases as it moves away from the nucleus?
 
@JohnRennie As I understand what you are saying is that in order to remove an electron from some energy state, you decrease it's kinetic energy which increases it's potential energy by moving away. So in practice in a lab, do you have an idea of how they decrease the kinetic energy of the electron?
 
@Alex No, you have the wrong end of the stick.
Suppose you wish to remove a stone from the surface of the Earth to infinity. The maths is the same as ionising a hydrogen atom. You give the stone some kinetic energy by throwing it upwards.
As the stone moves up the gravitational field of the Earth slows it down and reduces its kinetic energy. The total energy is constant so the reduction in KE is equal to the increase in PE.
If you throw the stone fast enough, i.e. increase its total energy to > 0, then the stone will head off into outer space.
 
@JohnRennie Oh yeah, I understand now :) Horrible confusion...But I see what I misunderstood now. Thanks.
 
In a hydrogen atom we generally give the electron increased kinetic energy by scattering a photon off it.
 
11:51 AM
@JohnRennie Okay thanks. That makes more sense than slowing it down to give it an escape velocity :)
 
12:40 PM
@Kaumudi.H: I've just bought another laptop. Should I seek medical help?
It's another fixer-upper. It should be easy to repair using bits I have lying around.
 
user228700
You're kidding, right?
 
user228700
@JohnRennie What are u gonna use that for?!
 
111
Hello everyone.
I was reading about Ampere's circuital law. My text says,
 
@Kaumudi.H Erm, good question :-)
 
111
'We choose the loop such that at each point of the loop, either :
1. B is tangential to loop.
2. B is normal to loop.
3. B vanishes. '
What do they mean by B vanishes?
 
12:49 PM
Actually I might keep it as a spare. It's the same as the one I use as a workstation i.e. the one I'm using to type this and it would be good to have a spare.
Unless you want another laptop :-)
 
user228700
A spare? Yep, seeking medical help would be the appropriate course of action.
 
Well laptops do go wrong ...
And if my main laptop went wrong I would be in trouble.
 
user228700
Are you telling me that you don't already have a spare? How did u resist the temptation for so long?!
 
But, well, basically you're correct and I should probably consult a psychiatrist :-)
5
Anyway it was cheap.
I could fix it and sell it for twice what I paid, though I won't of course.
 
user228700
...OK. If u buy another one in 6 months, I'm gonna have to insist that you consult a doctor about this.
 
12:53 PM
No, now I really don't have any excuses for buying more laptops
(though if you think that's going to stop me ... :-)
 
user228700
Yeah, no, that's what u said before.
 
Oh well :-)
 
111
@JohnRennie @Kaumudi.H Any clue about that thing?
 
@111 I'm not sure. If the magnetic field isn't constant there may be some locations where its value is zero.
If so putting parts of the loop at those locations makes the maths easy because obviously the flux at those points is zero.
 
111
Yeah, they're saying we do that for many practical applications. They're saying that we take the such a loop that it satisfies one of those 3, but how can a loop make B vanish?
Oops, the grammar.
 
1:02 PM
The loop doesn't make B vanish, but if B vanishes because that's the way the field is then you could arrange your loop to lie where the field is zero. I'd have to see the text you're referring to before i could say for sure what they are getting at. Is it a book?
 
111
Yes, but it is my syllabus book. Ever heard about the NCERT's ?
Wait, I'll type the whole passage.
'For several applications, a much simplified version of Ampere's Law proves sufficient. We shall assume that, in such cases, it is possible to choose the loop (called an amperian loop) such that at each point of the loop, either:
'We choose the loop such that at each point of the loop, either :
1. B is tangential to loop.
2. B is normal to loop.
3. B vanishes. '
Any example of a situation which illustrates the last point?
 
An obvious, if trivial, example is if the field is zero everywhere.
Apart from that nothing obviously springs to mind.
I'm not sure you should read too much into it. The book may be including the third option for completeness since it's technically possible.
 
111
Okay, so there
'might' be a situation where that happens, but usually it doesn't, and we can almost always choose loops which are parallel or normal to B. Right?
 
I think in real life things are rarely that easy :-) However for the purposes of the exam they will probably not give you situations that are too hard.
 
111
:D Alright. Thank you so, so much! :) Just one more thing and I'll stop pestering you. :P
 
1:13 PM
Yes?
 
111
Okay see, in a cyclotron, they say they put a source which oscillates the polarity of the dees. Can that source work on DC current?
 
The idea is that at the moment the particle leaves one of the Ds and starts crossing the gap between them, the other D is at a positive voltage so it attracts the electron and accelerates it across the gap.
So the voltage has to change sign every time the electron crosses the gap.
That's why it alternates. If it was a constant DC voltage the electron would just move ina circle of constant radius and wouldn't increase in energy.
Which would make it a pretty useless accelerator :-)
I have to head off now. Back in a couple of hours or so ...
 
111
Ok, thanks a lot! :)
 
¬¬
I was halfway through an answer for this and it got deleted.
 
111
Happens, @EmilioPisanty. :P
 
1:21 PM
@111 still. ¬¬.
 
@EmilioPisanty In the past when that has happened to me I posted the question and my answer as a self answered question. e.g.
3
14
Q: How does the Hubble parameter change with the age of the universe?

John RennieHow does the Hubble parameter change with the age of the universe? This question was posted recently, and I had almost finished writing an answer when the question was deleted. Since it's a shame to waste the effort here's the answer anyway. Maybe this can be one of the canonical answers sugges...

If you think it would be a useful contributiion then go for it.
 
111
XD John that is hilarious.
 
@EmilioPisanty Happens to the best of us :/
 
@JohnRennie I think that's what I'm going to do
 
@JohnRennie That's a really good idea, I'll keep that in mind
@JohnRennie Keep us up to date with your "broken computer" video series
First episode was great!
Autocorrect sucks...
Morning @ACuriousMind, or whoever just joined the chat. Mobile site isn't very good.
 
1:34 PM
@SirCumference mornin'
'Twas me indeed
 
 
1 hour later…
2:47 PM
@ACuriousMind Are you around?
 
@BernardoMeurer What's up?
 
Problems with a limit :P
$$\lim_{x\to0^+}{\sin{x}\log{x}}$$
Wait
I got it
 
3:02 PM
@BernardoMeurer L hospital ?
Or did you solve it algebraically ?
 
@anonymous Yeah, rewrite it as $\lim_{x\to0^+}\frac{\log x}{\frac{1}{\sin x}}$ apply L'Hopital and it works ok
 
@BernardoMeurer Good. Can you solve it without using L'Hopital ?
 
@anonymous Me? No. Someone who actually knows maths probably can though
I apply L'Hospital to everything
 
@BernardoMeurer :P
 
Just apply hospital
that's my rule of life
 
3:05 PM
that's the easy way :D

 SecretLabs (SE Branch)

The Labs, where the ideas are organised. Occassional attempt a...
What is this for ^ ?
@Secret Is that ^ your room ? Never saw it before!
 
@anonymous That's Secret's den
 
@BernardoMeurer I don't think secret is a wild animal in the jungle XD Haha :P
I can only see hundreds of questions from the main site on that room
 
3:29 PM
@ACuriousMind Teach me polynomial long division
 
Didn't you learn regular long division in high school?
 
@thoughtforfood I did yeah
 
It's the same setup
 
Okay, so say I have $$\frac{x^2}{x^2+1}$$, could you walk me through how you'd do it?
 
Khan academy is quicker :P
 
3:41 PM
@BernardoMeurer Uh, you can't do polynomial divison on that - the result of that thing is not a polynomial.
 
@ACuriousMind Crap
 
The long division algorithm terminates at the first step - the "solution" would be that the result is 0 with the term you started with as remainder.
 
So how the heck do I solve for $$\int\frac{x^2}{2}\frac{1}{x^2 + 1}$$
I did $\frac{1}{2}\int\frac{x^2}{x^2+1}$
And got stuck
 
You should read about partial fraction decomposition
 
@ACuriousMind Actually you can. The result of polynomial long division is that the degree of numerator must be lesser than the denominator. Though, you are right that there will be a remainder.
 
3:44 PM
@ACuriousMind I have, I have absolutely no clue how to do that
It's some chinese black magic BS
 
@BernardoMeurer Just write it as $x^2+1-1$. In numerator.
See what happens
 
You will get $\frac{1}{2}\int{1-\frac{1}{x^2+1}}dx$
 
You mean -, no?
 
@BernardoMeurer Ah, sorry, partial fractions don't help you here
 
3:47 PM
@BernardoMeurer Ah, yes :D
Now can you do it ?
 
@ACuriousMind The initial problem was solving for $$\int x\arctan x$$
So I integrated by parts
@anonymous Working on it :)
 
@BernardoMeurer The answer is $\frac{1}{2}(x-\tan^{-1}(x))+C$
Got it ? :D
 
$$\frac{1}{2}(x-\arctan x)$$
Yeah
 
Great!
Integration is fun :D
 
@anonymous Try solving $$\int x\arctan x$$
I got $$\frac{x^2}{2}\arctan{x} - \frac{1}{2}(x-\arctan{x})$$ see if you get the same :)
 
3:53 PM
@BernardoMeurer Absolutely correct!
 
Just add a C (The timeless joke) :P
 
I always leave that implicit here b/c I'm lazy
Everyone knows there's a C
 
It'll cost you marks.
 
3:54 PM
@BernardoMeurer Examiners don't know :D
 
That's why I'm not lazy doing exams :P
 
@BernardoMeurer Any more tough integrals you have ? Let me know when you find one :)
 
I'm solving one that isn't tough, but it's nice to solve, lemme type it in
Don't tell me the answer yet
 
Sure
 
$$\int_{0}^{\log{2}} \frac{1}{4-e^x}$$
 
3:58 PM
Looks like King's rule is needed here
 
I'm trying to substitute 4-e^x
 
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