The h Bar - 2017-01-28 (page 1 of 3)

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12:06 AM

@JohnRennie Particularly for special relativity right now (for an E&M course)

Bernardo Meurer

12:17 AM

@SirCumference That's even better actually

I need help testing something :P

1 hour later…

Bernardo Meurer

1:19 AM

@ACuriousMind

1

Q: Proving that a real function has maximum and minimum

Let $\psi : \mathbb R \to \mathbb R$ be a continuous function such that $\lim_{x\to+\infty}\psi(x) = +\infty$ and $\lim_{x\to-\infty}\psi(x) = -\infty$. If we define a function $G:\mathbb R \to \mathbb R$ such that $$G(x) = \frac{\psi(x)}{1+\psi^2(x)}$$ How can we evaluate whether $G$ has a maxi...

real-analysis proof-writing

Halp

@BernardoMeurer Have you considered what $G(x)$ does for $x\to\pm\infty$?

Bernardo Meurer

@ACuriousMind Yes, it goes to $0$ on both ends

But the function isn't necessarily differentiable

so I can't use Rolle

also, I gotta prove it has a max AND a min

@BernardoMeurer Correct - from above or below?

Bernardo Meurer

@ACuriousMind What do you mean?

@BernardoMeurer Is the function positive or negative as it goes to zero?

Bernardo Meurer

1:25 AM

On the left side it's negative, on the right positive

@ACuriousMind Ryan's writing a proof, we've been on this one for a while

@BernardoMeurer Right, so it has a zero somewhere

Bernardo Meurer

@ACuriousMind By the definition of a limit I guess so, yeah

By the definition of a limit? No, because the function is negative at some points and positive at other - a continuous function must have a zero to change sign

Bernardo Meurer

AH, I thought you meant on the endpoints

yeah, okay

Ah

I see what you're doing

go on

@BernardoMeurer No, they're limit to infinity and not in $\mathbb{R}$ - there are no endpoints

Bernardo Meurer

1:30 AM

@ACuriousMind I think in $\bar{\mathbb R}$

@ACuriousMind What Ryan is about to write you is absolutely true

@BernardoMeurer So call that zero $x_0$ with $G(x_0) = 0$. What can you say about $\min_{x\in [y,x_0]}G(x)$ and $\max_{x\in [x_0,z]}G(x)$?

Bernardo Meurer

@ACuriousMind Ryan asked me to link you this math.stackexchange.com/a/2117263/213493

@ACuriousMind Eh, I don't know what to say about those

@BernardoMeurer Well, that answer tells you already :P

Bernardo Meurer

@ACuriousMind Yep :)

Thanks though

@BernardoMeurer Ok?

Bernardo Meurer

1:42 AM

@SirCumference Can you use the command line?

@BernardoMeurer Depends

3 hours later…

thought for food

4:27 AM

0celo7 This user has been temporarily suspended by a moderator and cannot chat for 7 days.

3 hours later…

7:06 AM

hey guys, anybody now of a cheap polarized lightsource (for like DIY experimenting)

user image

@EmilioPisanty na i think this is more like it

And man, its still nice that I get this trickle of rep from back when i answered some questions really well. It makes me feel better about being lazy these days

2

@Skyler happens to everyone :-P

user228700

@JohnR: Morning :-)

Morning :-)

@JohnRennie What keeps the electron from falling down to the nucleus?

@Skyler d00d

I'm sorry I didn't get your calls.

user228700

7:20 AM

@DanielS: Greetings :-)

@DanielSank wat

@Kaumudi.H hi

1 sec

(getting phone now)

95

Q: Why don't electrons crash into the nuclei they "orbit"?

I'm having trouble understanding the simple "planetary" model of the atom that I'm being taught in my basic chemistry course. In particular, I can't see how a negatively charged electron can stay in "orbit" around a positively charged nucleus. Even if the electron actually orbits the nucleus...

quantum-mechanics electrons atoms models

@JohnRennie I saw this. This is asking from the perspective of the planetary model.

user228700

7:22 AM

@JohnRennie This question is what got me into physics back in 9th grade :-)

@Kaumudi.H wow

@DHMO Ben Crowell's answer explains why the uncertainty principle stops the electron collapsing inwards

user228700

@DHMO Hmm?

@Kaumudi.H interesting to see what got you here in the first place

it feels like

travelling back in time

user228700

...sort of :-)

7:25 AM

@Kaumudi.H I think that is true for almost every physics student (including me) :P The other question which boggled me is "Why do charges attract or repel each other?"

@JohnRennie but Ben Crowell's answer doesn't even have the word "uncertainty"

user228700

@anonymous Oh, cool! (I doubt if this is the case for "most" tho)

> when we confine a particle to a small space, we get a lower limit on its energy, and therefore once it's in the standing-wave pattern with that energy, it can't collapse; it's already in the state of lowest possible energy

@JohnRennie I don't understand the "therefore", and also the first sentence

@anonymous have you found the answer now?

@DHMO I don't think anyone has an answer to that :P Do you?

user228700

7:28 AM

@anonymous Whenever I began to ask such questions, a teacher I used to know used to say "This stuff is metaphysical. It's like asking why does gravity exist?"

I have an answer that goes into a bit more detail. Let me have a search and see if I can find it.

@anonymous how much do you want to know about quantum field theory?

74

A: Why do same/opposite electric charges repel/attract each other, respectively?

Well it has nothing to do with the Higgs, but it is due to some deep facts in special relativity and quantum mechanics that are known about. Unfortunately I don't know how to make the explanation really simple apart from relating some more basic facts. Maybe this will help you, maybe not, but thi...

@Kaumudi.H your teacher is wrong

and there is no problem with the example question your teacher gave

user228700

I agree that he could've put it better but after all, we weren't required to learn QFT.

@DHMO I wouldn't say that

7:30 AM

@DavidZ exploring the reasons behind the attraction of opposite charges is certainly scientific not metaphysical

@DHMO Thanks for that. I'm reading. But in most explanations I have seen, at the end we reach a point where we have to say "it happens because it happens" :P Anyway, I will read through that :)

@DavidZ I don't know much about QFT. Just started basic QM a few weeks back.

Hope to learn soon :)

@DHMO Not necessarily. It depends on the sense in which you mean "why". Science is first and foremost concerned with descriptions and predictions, and we can address reasons only to the extent that they emerge from the descriptive and predictive explanations we develop. Some people don't consider that sort of thing a valid answer to "why" questions.

@DavidZ by the "why" i really mean "mechanism".

@anonymous well I wouldn't get too excited, I spent 8 years learning QFT and I'm not sure I understand much of any of it even now

@DavidZ Scary :P I've heard JR say that same thing

7:35 AM

@anonymous no-one understands QFT - they just get good at faking it :-)

Something like that

While QFT works so well that it has to be correct (or a good approximation of correct) the perturbative treatment of interacting fields is just appalling. Either we are crap at maths or we've missed something somewhere.

@JohnRennie can you explain to me, in terms of QFT, why two electrons repel each other?

(I refrain from using proton because it is not fundamental)

No. Next question?

4

Why not?

7:39 AM

Because I don't know.

I thought you study QFT

Have you found your answer?

@JohnRennie So does QFT have any application in the real world or is it still only theoretical stuff ?

@JohnRennie next question: can I interchange + and -?

Precision tests of QED

Quantum electrodynamics (QED), a relativistic quantum field theory of electrodynamics, is among the most stringently tested theories in physics. The most precise and specific tests of QED consist of measurements of the electromagnetic fine structure constant, α, in various physical systems. Checking the consistency of such measurements tests the theory. Tests of a theory are normally carried out by comparing experimental results to theoretical predictions. In QED, there is some subtlety in this comparison, because theoretical predictions require as input an extremely precise value of α, which can...

I suppose not. A proton is made up of two positively charged quarks and one negatively charged quark.

7:42 AM

In physics, C-symmetry means the symmetry of physical laws under a charge-conjugation transformation. Electromagnetism, gravity and the strong interaction all obey C-symmetry, but weak interactions violate C-symmetry. == Charge reversal in electromagnetism == The laws of electromagnetism (both classical and quantum) are invariant under this transformation: if each charge q were to be replaced with a charge −q, and thus the directions of the electric and magnetic fields were reversed, the dynamics would preserve the same form. In the language of quantum field theory, charge conjugation transforms...

@JohnRennie but isn't a proton asymmetrical in terms of charge?

@DHMO C-symmetry means a proton and anti-proton behave identically, which they do. (Actually that's CP-symmetry but that's a detail :-)

I thought anti-proton travels backward in time

But, as you say, a proton is a composite object so it's more complicated than a fundamental particle.

@JohnRennie have you found it?

7:45 AM

@DHMO no, I'm afraid not.

@JohnRennie I must have made that up

53

Q: Is anti-matter matter going backwards in time?

Or: can it be proved that anti-matter definitely is nót matter going backwards in time? From wikipedia: There is considerable speculation as to why the observable universe is apparently almost entirely matter [...] the apparent asymmetry of matter and antimatter in the visible universe is on...

quantum-field-theory time antimatter

looks like it is a (popular) misconception

@DHMO no, you might have heard it said that an anti-proton is like a proton travelling back in time.

@JohnRennie is that trying to describe time symmetry?

@DHMO sort of. The symmetry that takes a particle to its antiparticle is CP-symmetry.

And the symmetry that sends a particle back in time is T-symmetry.

> if you apply the three operations of time reversal, charge conjugation (switch particles and antiparticles), and parity inversion (mirroring space), the result should be exactly equivalent to what you started with.

I think classical mechanics is t-symmetrical

7:47 AM

The combination of the two is CPT-symmetry, and we believe all particles are invariant under CPT-symmetry.

Charge, Parity, and Time Reversal Symmetry is a fundamental symmetry of physical laws under the simultaneous transformations of charge conjugation (C), parity transformation (P), and time reversal (T). CPT is the only combination of C, P, and T that is observed to be an exact symmetry of nature at the fundamental level. The CPT theorem says that CPT symmetry holds for all physical phenomena, or more precisely, that any Lorentz invariant local quantum field theory with a Hermitian Hamiltonian must have CPT symmetry. == History == The CPT theorem appeared for the first time, implicitly, in the work...

So if you apply CP to convert a particle to its antiparticle, then T to send it back in time you should get the original particle back.

So, what keeps electrons from falling?

So a proton is equivalent to an anti-proton moving back in time, and vice versa

I see

7:49 AM

@DHMO OK, I can't find my answer but I can give you a summary here.

@JohnRennie ok

you have 4,723 posts...

and 20 questions...

that is not very... symmetrical.

When we say an electron is confined we mean its wavefunction goes to zero at infinity.

agreed

So the wavefunction starts at zero at infinity, increases to a non-zero value at the (average) position of the particle and goes to zero again as move away on the other side.

@JohnRennie why on earth did you put all but one question in community wiki?

7:52 AM

That means both the first and second derivatives $d\psi/dx$ and $d^2\psi/dx^2$ must be non-zero.

agreed

user228700

@JohnR: Do get back to this please, when you have finished this discussion but my latpop charger overheats sometimes, is that OK?

@Kaumudi.H get back to what?

user228700

@DHMO To my question about my charger.

@DHMO If we squeeze the particle inwards then the first and second derivatives increase because they have a shorter distance to increase in.

7:54 AM

@JohnRennie not agreed... is $\displaystyle \int_{\text{space}} \psi \ \mathrm dx$ fixed?

The wavefunction is normalised so $\int\psi^*\psi\,dV=1$

ok, agreed

That's just saying that the probability of finding the particle somewhere must be unity.

I thought wavefunction is not probability

$\psi^*\psi dV$ is the probability of finding the particle in the volume element $dV$

7:58 AM

please continue

Are happy with the claim that $d^2\psi/dx^2$ gets bigger (in magnitude) as we confine the particle inwards?

I suppose so

This is the key point, so you need to be happy about it otherwise the rest of the argument won't convince you.

@Kaumudi.H: does the charger just get warm, or does it do something more spectacular?

user228700

I didn't notice until today but it would be something b/w spectacularly hot and warm.

The chargers do get warm, especially if the battery is very low so they have to charge at full power for a while.

user228700

8:03 AM

...that is what just happened.

@JohnRennie is it difficult to prove?

@DHMO I have to say it seems intuitively obvious to me. The wavefunction starts at zero with zero derivatives and has to rise to a finite value and then fall back to zero ina finite distance.

As you reduce that distance it has to change more rapidly so the gradient has to be greater.

ok, agreed

And for the gradient to start and zero and reach a greater value in a shorter distance the second derivative has to increase as well.

@Kaumudi.H It got warm while charging a low battery?

user228700

...yes.

8:06 AM

@Kaumudi.H if so, as long as it doesn't actually start smoking I wouldn't worry.

user228700

And I mean the rectangular box in the middle.

user228700

@JohnRennie OK, good to know (:-P), thanks.

Yes, that's the charger. The box with a cable that plugs into the mains at one end and a cable that plugs into the PC at the other.

Does it actually get too hot to hold?

user228700

Well, I dunno what u mean by that but it's not scalding hot, no.

In that case it's fine.

user228700

8:09 AM

OK. Good to know :-P

user228700

How's ur Saturday going?

There is another laptop just like yours on eBay right now. I'm getting twitchy :-)

user228700

Don't do it.

But, but, but ... :-)

user228700

You're not going to have any use for it, no?

user228700

8:10 AM

(Well, of course, you could invent some use for it)

@Kaumudi.H I would never do such a thing. Actually I need a door stop to hold my living room door open ...

user228700

Ah, yes, and a laptop could serve as just that, after you're done playing with it of course, yes? :-)

user228700

Also, why are u looking to hold your living door open? Isn't it freezing out?

I need a spare laptop for my niece. If she breaks hers (entirely possible since she's 15) she'll be on the phone demanding I fix it. There! That's an excuse,err, I mean reason to buy another laptop :-)

You've gone quiet. I can see you sitting there shaking your head sadly :-)

@JohnRennie Or a bunch of thieves could attack your house tonight and steal all your laptops :P. So you should order one just in case... :P Or maybe an earthquake.....Or tornado.... :P So many reasons :D

8:15 AM

@anonymous yes, of course, thanks :-)

@JohnRennie could we continue?

@DHMO oops, yes, sorry. Actually we're basically there because the second derivative is related to the energy by the Schrodinger equation:

$$ \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi$$

So confining the particle increases its energy, and confining it to a point sends the energy to infinity.

eh, what?

-1

Q: how can we distribute charge on charged concentric spherical shells when they are connected through conduting wire?

please help me sooonnjjjjjjlk;blIAhsipaenr;akljfpo ahfENQ;OPFJ;DFLWE:oJP;NHDIWPYIPRHyrfepihf;k

Idiot. Why we should take his question seriously escapes me.

Since he clearly doesn't.

@JohnRennie could you explain that equation?

8:22 AM

@DHMO ignore the potential for now, then: $$\frac{d^2\psi}{dx^2} \propto E\psi $$

@DHMO Explain the Schrodinger equation?

@JohnRennie ya, pretty much

No, I can't explain the Schrodinger equation in a chat room. You'll need to either just accept it or read up on it. You're basically asking me to explain all of quantum mechanics.

People write 500 page textbooks to do that.

ok, I accept it

now tell me why that does not apply on the macro scale

14

A: Validity of naively computing the de Broglie wavelength of a macroscopic object

If you've read about optical diffraction experiments like the Young's slits, you may have noticed they all refer to coherent light. This is the requirement that all the light in the experiment is in phase. If you aren't using coherent light you won't observe any diffraction because different bits...

so you're saying that i'm naive

8:28 AM

I'm saying only that you don't know quantum mechanics. Neither did I when I was your age - it isn't a crime :-)

You only start learning this stuff in your first year at university.

I only saw words in your answer

user228700

@JohnRennie Oh, sorry, there are guests at home.

user228700

But yes, I am shaking my head sadly :-P

@JohnRennie i'm not planning to study physics lol

@Kaumudi.H :-) I am just joking. I might be tempted if the laptop was very, very cheap, but it isn't. Especially since I bought a (very, very cheap) spare laptop just the other day I really have no use for another laptop.

@DHMO if you do chemistry I'm afraid you'll still need to learn QM :-)

user228700

8:32 AM

Breathes a sigh of relief--I thought u were becoming loony.

@DHMO What are you planning to study ?

user228700

Maths?

@DHMO I say this because I did chemistry (physical chemistry) not physics.

Even some mathematicians learn QM I think

@JohnRennie @anonymous I'm planning to do maths

8:40 AM

Is this a good book ? amazon.com/dp/0306447908/?tag=stackoverfl08-20 Anyone here used it ?

user228700

Lol x'D

@anonymous Shankar's book seems to have mixed reviews

@JohnRennie I didn't imagine so much discussion had already taken place on the book :P

I think I should give it a try :)

user228700

@anonymous That's why the "Lol x'D"

Last night dream ( h bar section):

In h bar, I made an announcement that 0celo7 is back. This is later starred 3 times. A user (mix of Doc x Anonymous) wrote excitedly all in caps saying really? (and 4 other sentences I forgot). Shortly after 0celo7 commented. There's also a 0celo3 commented that he is a more mature version of his past self. There's a post than later get starred 26 times.

8:54 AM

@Secret so you want 0celo7 back desparately

user228700

@Secret 26? That's likely :-P

@JohnRennie do you believe that travelling faster than light brings us back in time?

DHMO: pbs.org/wgbh/nova/blogs/physics/2015/08/…

It's more complicated than that

@DHMO I don't believe that travelling faster than light is possible. If it were possible then you can show that FTL travel could be used to construct a closed timelike curve.

@JohnRennie I suppose travelling faster than life brings you to death

4

8:57 AM

:-)

@DHMO I originally intended to do maths. In fact I was offered a place to do maths at Imperial College but decided to wait a year and apply to Cambridge instead. Just as well really as I wouldn't have made a good mathematician.

@JohnRennie i'm in fact waiting for an offer to do maths at Imperial College...

Cool :-)

@DHMO You live in London ?

@anonymous no I don't

so if I make it

the tuition fee would be

how do you say

@DHMO transfinite? :-)

9:06 AM

@JohnRennie yes

Won't they give fee waivers ?

But then Imperial College is one of the top universities in the UK. My brother went to Imperial College to study biology.

@JohnRennie Did the hajmola (which you ordered) reach you ? :P

@anonymous Indeed yes. What surprised me is how salty they are. The flavour is quite nice but they are so salty I find them hard to eat. The tamarind ones are not too bad, and I still eat them from time to time, however the original flavour ones have gone in the bin.

I expected that :) Western cuisine is definitely not as spicy as in Asia and probably that is the main reason. That's why I recommended buying the imli (tamarind) one rather than the original one :D

9:20 AM

It's not the spiciness - I like chillis and eat them regularly. The Hajmola are just so salty.

Anyhow I have to work for a bit. Back in an hour or so ...

thought for food

Cya

Biyyaaa! :)

I am going to test something

is there a limit to this nesting?

@Secret Go on testing :P

Maybe there isn't any limit

Unless the server memory is over (which is unlikely)

9:48 AM

@Secret I broke the limit

your sentence is no longer showing

(it is still showing in the page source so...)

Anyone interested in solving a challenging indefinite integral ?

Try this:

@Secret never mind

@anonymous I'm up

$$\int({\sin(x)})^{\frac{-11}{3}}({\cos(x)})^{\frac{-1}{3}} dx$$

Try it :D!

$=\displaystyle\int(\sin x)^{-4}(\tan x)^{1/3}\ \mathrm dx$

I'm inclined to use Weierstrass substitution

(and end up with a degree-12 rational function)

Actually its a very easy problem in disguise. Just 2 or 3 steps.

I felt like kicking myself when I came to know the solution.

No Weierstrass needed

9:54 AM

let me see...

$\dfrac{\mathrm d}{\mathrm dx} (\sin x)^{1/3} = \dfrac13 (\sin x)^{-2/3} \cos(x)$

no

$\dfrac{\mathrm d}{\mathrm dx} (\cos x)^{2/3} = -\dfrac23 (\cos x)^{-1/3} \sin(x)$

Then $u=(\cos x)^{2/3}$

no it doesn't seem right

Hint: You need to use derivative of $tan(x)$ somewhere :)

hmm...

@Secret Are you trying the integration? :)

$I=\displaystyle-\frac32\int(\sin x)^{-14/3}\ \mathrm d(\cos x)^{2/3}$

$I=\displaystyle-\frac32\int\frac1{(1-u^2)^7}\ \mathrm du$

@anonymous I use whatever I want

Currently a bit busy in some time travelling stuff. I will deal with that later

10:01 AM

@DHMO Seems close. But what next ?

@anonymous partial fraction...

@DHMO Err...toooo big

That is correct though

alright, let me try...

$I=\displaystyle\int(\tan x)^{-11/3}(\cos x)^{-4}\ \mathrm dx$

yes I'm getting there

$I=\displaystyle\int(\tan x)^{-11/3}(\sec^2 x)^2\ \mathrm dx$

Yes yes yes

$I=\displaystyle\int t^{-11/3}(t^2+1)\ \mathrm dt$ where $t=\tan x$

10:05 AM

Yes you got it. Great !

$I=-\dfrac{3}{2t^{2/3}}-\dfrac{3}{8t^{8/3}}+C$

$I=-\dfrac{3}{2(\tan x)^{2/3}}-\dfrac{3}{8(\tan x)^{8/3}}+C$

thanks for your exercise @anonymous

Well done. I have one more for you if you are interested :)

@DHMO

yes I am interested

Okay. Lemme type it

user image

@DHMO Here you go

This is also a easy one in disguise :)

that's an interesting way of typing...

$\displaystyle \int \left[ \frac {\sqrt{x^2+1} \left[ \ln(x^2+1)-2\ln x \right]} {x^4} \right] \ \mathrm dx$

10:10 AM

@DHMO Ya :P

I swear, $u=1+\dfrac 1 {x^2}$

Close :)

Then I proclaim that $\mathrm du = -\dfrac2{x^3}\ \mathrm dx$

Yes you did it. Great! Let me find a tougher problem for you :).

but you said close meaning that I'm wrong?

10:14 AM

@DHMO No. You identified the crux of the problem! Close means "near"

Then $I=\displaystyle-\frac12\int\sqrt{u}\ln u\ \mathrm du$

@anonymous interesting use of the word "close"

@anonymous expecting...

Then I announce that $t=\sqrt u$ and $\mathrm du = 2t\ \mathrm dt$

@JohnRennie welcome back

i.stack.imgur.com/amxTG.jpg I found this a bit complicated at first. You can check it out :D

$I=-2\displaystyle\int t^2\ln t\ \mathrm dt$

@anonymous interesting

→ 16 messages moved to Trash

user image

10:18 AM

→ 1 message moved to Trash

@Secret please don't do that in the main chat room

ok

$\displaystyle\int\frac{x^2+x}{(e^x+x+1)^2}\ \mathrm dx$

^ It needs some manipulation of $e^x$

@anonymous you see, it gets messy when you mix $e^x$ and $x$ together...

@DHMO Need a hint ? It is once again a 2 step problem :)

10:23 AM

you remind me again of synthesis lol

and no, I don't need a hint

Hehe. Try taking $e^x$ common out from the denominator and see the derivative of what remains inside the bracket.

$\displaystyle\frac{\mathrm d}{\mathrm dx}(1+xe^{-x}+e^{-x})=-xe^{-x}$

Ya. Close again. Now substitute the denominator with $y$ and see the magic!

$I=\displaystyle-\int\frac{(x+1)e^{-x}}{y^2}\ \mathrm dy$

Write the numerator as $y-1$

10:28 AM

$I=\displaystyle-\int\frac{y-1}{y^2}\ \mathrm dy$

Yesh!

this is witchcraft

Indeed!

user image

Next One!

can I do $\displaystyle\int\frac{\mathrm dx}{\sin x\sqrt{\cos^3 x}}$ instead?

yes you can

try it

10:31 AM

$u=\sqrt{\cos x}$

$2u\ \mathrm du = -\sin x\ \mathrm dx$

I'm coming back in 5 mins

Hi, everybody.

$I=\displaystyle-\int\frac{2\ \mathrm du}{(1-u^4)u^2}$

@DanielSank hi

Are we playing "solve the integral"?

yes

10:32 AM

Hmm, that can be a fun game.

$t=\sqrt{\tan x}$

$2t\ \mathrm dt=\sec^2x\ \mathrm dx$

this isn't working

@anonymous

@DHMO This looks good. Will you use partial fraction after this ?

do you mind if I repost the previous problem with $e^x$ to another chatroom?

@anonymous I can but I'm seeking better solutions

@DHMO Haha. That problem isn't copyrighted. You are free to share :)

$y=\dfrac1u$

$\mathrm dy = -\dfrac1{u^2}\ \mathrm du$

$I=\displaystyle\int\frac{2y^4\ \mathrm dy}{y^4-1}$

I can use partial fraction on this @anonymous

what is your solution?

10:45 AM

@DHMO I think it can be written as $$\frac{\sec^2(x)}{\tan^{\frac{1}{2}}(x)}$$

@anonymous I don't think so

Oh no sorry

$\dfrac{\sec^2 x}{\tan^{1/2}x} = \dfrac{1}{\sqrt{\sin x\cos^3x}}$

I missed the sine at the bottom

ya I thought of something similar

10:47 AM

@DHMO Yes, I solved it like that only

I don't think there is any shortcut

@anonymous this is the most beautiful proof I've seen

apart from Gaussian integral

@DHMO I think it would be easier with complex numbers using Euler's identity

@anonymous what do you mean?

Use $e^{ix}=\cos(x)+i\sin(x)$ to substitute the sine term

Then use by parts

It could be another method of the doing the proof

I haven't tried it though

How do you integrate $\int \frac{e^x}x\ \mathrm dx$?

10:53 AM

Hi.

15

Q: Problem when integrating $e^x / x$.

I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong? So, I'm trying to determin...

calculus sequences-and-series integration fake-proofs

It doesn't seem to have an indefinite integral though

exactly

Okay so not a very good idea :)

The proof you linked is better]

Hii @DHMO

hi

10:58 AM

user image

Could you help me in a physics question @DHMO

not if you don't ask it

This one is also a good one ^

Here it is

user image

in that first we got i=9.1 and after that we got i=1A how ?

@anonymous do you want a deal? you help koolmand and I do your integral

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