Mathematics - 2013-08-30 (page 1 of 2)

eXtremiity

1:24 AM

Hi there, can someone teach me how to find where a complex is differentiable?

Here's an example f(z)=z|z|. I understand that I have to use a theorem. I'm just unsure of how to implement it.

1) C-R Equations must satisfy
2) Partial derivatives of C-R must be continuous.

How can I combine the theorem with actually finding the -points- of differentiability.

eXtremiity

1:35 AM

I can't seem to find any points where the complex function f(z)=z|z| is differentiable.

eXtremiity

1:46 AM

Well, what I have gathered that while the C-R equations tell us where the function is differentiable - IF the C-R equations do not exist at a specific point, then it does not necessarily mean that the function is differentiable at that point.

Peter Tamaroff

2:44 AM

@TedShifrin Le Ted.

@robjohn Le Robbie.

robjohn

@PeterTamaroff >8(

Peter Tamaroff

@robjohn Oh, noes. I startled the square.

eXtremiity

whats the process in determining where a complex function is -analytic- ?

Kevin Driscoll

A complex function is analytic at a point if it can be expressed as a power series in some small neighborhood of that point @eXtremiity

Which I believe requires that the function and all of its derivatives be well defined and continuous at that point

eXtremiity

So, since I've heard other definitions - I am taking that one you have described to me, is one of them.

There's also the neighbourhood definition.

Kevin Driscoll

2:59 AM

@eXtremiity As far as I know they are equivalent

eXtremiity

I'm afraid I can't see that. But I guess it's not important right now. Could you assist me with an example?

$e^z (z-\bar{z})^2$

Kevin Driscoll

@eXtremiity basically to say that $f$ is analytic at $z$ you must prove that its Taylor Series converges at $z$

eXtremiity

Ok. That makes sense.

So to find where the function is analytic, I would create a taylor series and see what values of z will cause it to converge?

Kevin Driscoll

Exactly

eXtremiity

I see. Is there any need to go into the Cauchy Riemann equations?

Just asking because I finished learning differentiability of a complex function.

Kevin Driscoll

3:06 AM

To determine analyticity? I don't think so

I mean strictly speaking the C-R equations must be satisfied for a function to be complex differentiable

but for something as benign as the example you posed, I don't think there is a problem

eXtremiity

Yes, for the example I provided:
$e^z (z-\bar{z})^2$
It simplifies to $-y^2(e^{x+iy})$

I'll use the taylor series for the exponential function, and I'll get something nice (relatively).

$-y^2(e^{z})$

Now the trouble I'm having is seeing where the taylor series will converge for its complex nature.

Kevin Driscoll

So z and y are not independent

you can't carry the derivative through y

eXtremiity

What do you mean by your second line?

You can't carry...

Kevin Driscoll

I dean that $\frac{d y^2 e^z}{d z} \neq y^2 \frac{d e^z}{d z}$

After looking at the example, because it involves complex conjugation I think it actually is easier to use the CR equations here

Complex conjugation is tricky, you don't want to assume that it is differentiable

eXtremiity

I found that it is differentiable where y=0.

using CR equations.

Kevin Driscoll

3:19 AM

Did you check that BOTH the CR equations are true simultaneously?

oh yes I see

eXtremiity

Yes I did. Are you suggesting that I did something wrong?

Kevin Driscoll

I think that is the only correct answer yes

eXtremiity

Yeh, I'm sure I did.

Kevin Driscoll

for a moment I thought there were more points, but I checked carefully and I was wrong

eXtremiity

Oh, and the answers say so too. So we don't have to worry :) .

Just the analytic part. Yeh, I know where you're coming from. The sin and cos made me believe that there'd be more points. But they can't both hold simultaneously, so y=0 is the only solution (given that you've seen the CR equations).

Kevin Driscoll

3:20 AM

So surprise surprise, the function is differentiable only on the real line

eXtremiity

Yup!

Kevin Driscoll

and you know what the offending part is because $e^z$ is entire, ie analytic everywhere

so its the other part that is preventing the function from being differentiable

eXtremiity

e^z is hanging around with the wrong functions.

Kevin Driscoll

accesory after the fact, he might get probation

eXtremiity

Haha xD

So where were we?

Yes, so by CR we know the function is differentiable at y=0 -> only on the real line. How do we relate that to where the function is analytic?

Kevin Driscoll

3:24 AM

Well if the function is differentiable

at y=0

Then if you plug in y=0

you get a function that is 0 everywhere in x

eXtremiity

independent of x. yep.

Kevin Driscoll

Indeed, and the 0 function is a perfectly good Taylor series

so the function is analytic everywhere on the real line

eXtremiity

Hm, well the answers say that it's analytic nowhere. But the answers, have been wrong the last few questions .

Kevin Driscoll

AH no actually I made a mistake

the book is right here

You need a stronger condition for the Taylor Series to converge than that the CR equaitons are satisfied at the point

Peter Tamaroff

@anon Dawg.

Kevin Driscoll

3:29 AM

You need that the function is complex differentiable in a neighborhood of the point

and in this case this is not true, the function only satisfies CR equations on the real line

but if you pick any point on the real line, no matter how small a circle you draw around it you will always be including point where the CR equations are violated, so the function is not holomorphic on the real line

errr analytic I mean

eXtremiity

Could you perhaps pictorially describe to me :
"You need that the function is complex differentiable in a neighborhood of the point"

Because, whenever I hear someone say it - I think of a ball with a line (the function) going through it.

And that all parts of the line -inside- the ball have to be differentiable. Am I right =/ ?

Kevin Driscoll

What it means is that you pick some point $z_0$

a neighborhood of $z_0$ is just a little circle in the complex plane with its center at $z_0$ and then you make the radius of the circle arbitrarily small

Peter Tamaroff

@eXtremiity A function is said to be analytic in an open subset $S$ of $\Bbb C$ if it admits a derivative there. Apparently one can prove that this derivative, if it exists, must be continuous. Quite dopey.

eXtremiity

Hmmm ok.

Karl Kronenfeld

@eXtremiity The graph of the function is 4-dimensional. But you can graph only the real part or the imaginary part as a surface in 3-D space.

Kevin Driscoll

3:34 AM

I don't think you can think of a complex function as a curve though. A complex function has a complex value everywhere in the plane.

eXtremiity

@KevinDriscoll . I think I get it :) .

Kevin Driscoll

What Karl said.

eXtremiity

Oh I see.

So we can't create a epsilon disk small enough so that all the points inside the disk are differentiable.

Kevin Driscoll

Right, exactly

eXtremiity

Ahh, I get it. Makes sense.

So thats for "this" type of question.

How would I -approach- other questions?

Taylor series still a good method?

Kevin Driscoll

3:36 AM

If you are confident that the Taylor series exists then you can jump straight to finding the Taylor series

but if you are unsure how to take the Taylor series or if the function is differentiable anywhere, best to check the CR equations first

eXtremiity

So something else I can take from this.....is that if a function is only differentiable on a particular line on the complex plane - in it's never going to be analytic ?

Kevin Driscoll

So, for something like $\frac{\sin(z)}{z-i}$, I know that polynomials and the sine are well-defined in terms of their taylor series in the complex plane so I'd just straight to looking at the series.

eXtremiity

I see.

Kevin Driscoll

Yes, thats correcy

correct*

eXtremiity

Excellent. Well @KevinDriscoll , I can't thank you enough for your patience mate

I'm going to have a break now - I've passed my 1 hour study sesh, need to have a break so I can recharge and get back to it.

Thanks again man !

Kevin Driscoll

3:40 AM

No problem. I am relearning all my complex analysis myself right now

eXtremiity

Awesome . :]

Peter Tamaroff

3:52 AM

@eXtremiity One can give the following insight. By the Cauchy Riemann equations, we know that $$Df(z_0)=\begin{pmatrix} D_1f_1(z_0)&D_2f_1(z_0)\\-D_2f_1(z_0)&D_1f_1(z_0)\end{pmatrix}$$ so $Df(z_0)$ is multiplcation by $f'(z_0)$.

A complex function is analytic when it locally simply rotates and expands the space, namely by a factor of $|f'(z_0)|$ and an angle of $\arg f'(z_0)$.

@eXtremiity You can probably enjoy Needham's Visual Complex Analysis.

skullpatrol

4:11 AM

:)

Kevin Driscoll

@PeterTamaroff Do you know what we can say about the difference between the Riemann integral of $f(x)$ and the Riemann integral of $\Sigma f_n(x)$ where the $f_n$ converge to $f(x)$ except at a single point?

Peter Tamaroff

@KevinDriscoll Converge how? Pointwise? Uniformly?

Kevin Driscoll

Oh right

umm let me look up the difference

Peter Tamaroff

I can tell you the difference.

You needn't look it up.

@KevinDriscoll

Kevin Driscoll

I just read the wiki page

Peter Tamaroff

4:15 AM

OK.

The thing is this.

If $f_n\to f$ poinwisely, there is no certainty that $\int_I \lim f_n=\lim \int_I f_n$

Kevin Driscoll

Okay

I suspect the convergence is pointwise, but am not positive

Peter Tamaroff

Well, details?

Kevin Driscoll

I was specifically interested in the asymptotic series expansion of $\text{sech}(x)$ for large $x$

ie $$\text{Sech}(x) = 2\sum_{n=0}^{\infty} e^{(2n+1)\frac{\pi x}{2}}$$

Peter Tamaroff

@KevinDriscoll In the name of all that's holy, how is that convergent?

skullpatrol

Hi @unknown now that you can chat, you don't have to be unknown :D

Peter Tamaroff

4:20 AM

Also, use \sum, not \sigma.

As in $$\sum_{n\geqslant 0}$$

Kevin Driscoll

uuuuuuum, I believe that it converges everywhere except $x=0$

Ah its lowercase 's' I kept trying '\Sum' and it wouldnt work

I mean for positive x of course

OH TYPO

Karl Kronenfeld

You're missing a minus sign, if I had to guess.

Kevin Driscoll

Bingo @KarlKronenfeld

Peter Tamaroff

I find it more aesthetically pleasing to write it as $$2\cdot \sum_{n\geqslant 0}\exp\left(-\left(n+\frac 1 2\right)\pi x\right)$$

And what is it that you want?

Kevin Driscoll

Indeed. But is it uniformly convergent?

Peter Tamaroff

4:25 AM

@KevinDriscoll Yes. We can dominate it by a convergent sum, methinks.

Wait, what is "Sech" here?

Kevin Driscoll

The multiplicative inverse of "Cosh"

Peter Tamaroff

But you have $\pi x$.

Kevin Driscoll

Oh I am stupid, please excuse my carelessness

Peter Tamaroff

Also, I think you want an alternating sum.

But to make it short, you can integrate this termwise, yes.

Kevin Driscoll

$$\text{Sech}(\frac{\pi}{2} x) =2 \sum_{n=0}^{\infty} (-1)^n e^{-(n+\frac{1}{2})\pi x}$$

Peter Tamaroff

4:31 AM

@KevinDriscoll Yes, that looks good now.

Kevin Driscoll

I apologize it has been a long day

Peter Tamaroff

@KevinDriscoll No need to do that.

Kevin Driscoll

But yes, term by term integration doesn't seem to be a problem

Peter Tamaroff

An example and another.

Kevin Driscoll

Ah right of course

Peter Tamaroff

4:34 AM

@KevinDriscoll If I were you, I'd stick to Lebesgue integration.

skullpatrol

Hi! @PSTikZ long time no see?

Kevin Driscoll

Heh. I wish I could but I don't know anything about Lebesgue integration. I know its 'nicer' but physicists never teach it

PSTikZ

@skullpatrol hi baby :-)

Kevin Driscoll

@PeterTamaroff

Peter Tamaroff

@KevinDriscoll There is a theorem that says that if $g_n\geqslant 0$ on $I$; $\sum_{n\geqslant 0}g_n$ converges (pointwisely) a.e. on $I$ to some function bounded above by a Lebesgue integrable function then $$\sum_{n \geqslant 0}\int_I g_n$$ converges and $$\int_I \sum_{n\geqslant 0} g_n=\sum_{n\geqslant 0} \int_I g_n$$

skullpatrol

4:37 AM

@PSTikZ How are you honey?

Peter Tamaroff

@KevinDriscoll You can always read about it.

Kevin Driscoll

AH okay, is this the dominated convergence theorem? @PeterTamaroff

Peter Tamaroff

@skullpatrol You two, get a room.

PSTikZ

@skullpatrol Fine, well-balanced. And you honey?

Peter Tamaroff

@KevinDriscoll Nope, this is kind of a corollary to DCT.

skullpatrol

4:38 AM

@PSTikZ Fine thanks sweetie ;-)

PSTikZ

@skullpatrol :-)

Kevin Driscoll

I have read about it briefly but honestly I lack the background to understand it. I don't speak the language of pure mathematics. @PeterTamaroff

Peter Tamaroff

@KevinDriscoll Recall DCT is $f_n$ converges a.e. on $I$ to some $f$ and $|f_n|\leq g$ for some Lebesgue integrable $g$ on $I$, then the sequence $\int_I f_n$ converges and $\int_I f=\lim \int_I f_n$.

@KevinDriscoll Hmm. There is a measure theoretic free approach in Apostol's book. I am thinking you can read that. Dunno if people like it or not.

Kevin Driscoll

@PeterTamaroff Might be worth looking at, But yes the measure theory business I just can't parse yet. I understand what $dx$ is physically. $d\mu$ I have no idea

Sadly, it seems that to compute what I am looking for I would have to take the inverse of this sequence, multiply it by another function and integrate. I was hoping that because the sequence is simple this operation might generate an integral with a known result. This seems to not be the case though; the integral is not known by Mathematica/

I actually should have guessed this. Based on the structure of what I was doing there is no way to integrate 'term by term' because the sequence is in the denominator. "DOH!"

Peter Tamaroff

@KevinDriscoll The inverse of what?

Kevin Driscoll

4:45 AM

Sorry by inverse of the sequence I mean just 1 over the asymptotic series

But I learned something about pointwise vs uniform convergence so a net positive I'll say

Peter Tamaroff

@KevinDriscoll Err, but why not just work with $\cosh x$? :confus:

skullpatrol

Yo @GustavoBandeira wazzup?

Kevin Driscoll

@PeterTamaroff In what I am looking at the expression is $1 - \frac{1}{s \text{Cosh}(\pi \frac{x}{2})}$ which I then have to reciprocate to put on the other side of an equation and integrate

Gustavo Bandeira

@skullpatrol Hello!

@skullpatrol Not much.

Peter Tamaroff

@KevinDriscoll Use \cosh

Kevin Driscoll

4:48 AM

I don't know why I though writing out the asymptotic series would be simpler

in hindsight it seems silly

Gustavo Bandeira

@skullpatrol You?

skullpatrol

@GustavoBandeira Chillin' like a villain.

Peter Tamaroff

@KevinDriscoll Ah, OK. You have to invert that in a nbhd of $0$? That is nasty business.

Oh, wait.

You cannot invert that in a nbhd of $0$.

Kevin Driscoll

@PeterTamaroff AH sorry me being silly again, should be Sinh and not Cosh

Peter Tamaroff

That blows up at the origin. OK.

And $s$ is a free variable?

Kevin Driscoll

4:51 AM

@PeterTamaroff Indeed it does. Yes.

Peter Tamaroff

@KevinDriscoll Well, I can give you this.

Kevin Driscoll

@PeterTamaroff You can see the integral equation I have been trying to gain some analytical understanding of the solution of since April near the bottom here: math.stackexchange.com/questions/346924/…

Peter Tamaroff

You can "just" write $$\frac{1}{{1 - \frac{s}{{\sinh x}}}} = 1 + \frac{s}{{\sinh x}} + \frac{{{s^2}}}{{{{\sinh }^2}x}} + \cdots $$

@KevinDriscoll That beats the crap out of me.

I have to sleep now. Ugh. Meh brainz are too tired.

Kevin Driscoll

@PeterTamaroff yes, I don't expect anyone knows how ot solve it. Too complicated.

@PeterTamaroff That would converge for small s, yes?

Peter Tamaroff

@KevinDriscoll If $x$ is fixed, for $|s|<|\sinh x|$.

Kevin Driscoll

4:56 AM

@PeterTamaroff Makes sense, thanks for your time. Have a good rest

Peter Tamaroff

Is $$f(s) = \frac{1}{{a\beta }}\int_{ - \infty }^\infty f (s')\frac{{s\sinh \frac{{\pi s}}{2}}}{{s\sinh \frac{{\pi s}}{2} - 1}}\frac{1}{{\cosh \frac{{\pi s'}}{2}}}ds',$$ supposed to read

$$f(s) = \frac{1}{{a\beta }}\int_{ - \infty }^\infty f (s')\frac{{s\sinh \frac{{\pi s'}}{2}}}{{s\sinh \frac{{\pi s'}}{2} - 1}}\frac{1}{{\cosh \frac{{\pi s'}}{2}}}ds',$$

Ah, no.

OK.

Kevin Driscoll

No, it is correct as written.

Peter Tamaroff

Misread.

The why don't you pull that out the integral?

Kevin Driscoll

I could. I did not in this writing because the form I have it in is 'canonical' for the field

Peter Tamaroff

5:09 AM

@KevinDriscoll Ah. Now I go.

skullpatrol

Oops sorry

Karl Kronenfeld

@skullpatrol I believe he has done that. Moreover, I think it is the content of the last link he posted.

skullpatrol

maybe Physics.SE?

Kevin Driscoll

Precisely. @KarlKronenfeld @skullpatrol From my reading and discussions since then I gather that this is (1) In the general case not a tractable problem and (2) The form of integral equation that I posted where the poles are not 'on the diagonal' (in this case when s=s') is non-standard.

@skullpatrol I've thought about it, but I don't think they would be very helpful. Most physicists don't deal with integral equations.

It is just a mess. None of the standard approaches work.

I suppose that is why it is a research-level problem and not in a textbook :-P

skullpatrol

Put it away for awhile and come back to it later...

...maybe MathOverflow.SE

Kevin Driscoll

5:20 AM

@skullpatrol I have a numerical solution and a paper forthcoming (if my advisor decides to finish it). I have considered MathOverflow but I don't know that they are interested in such problems that are really more Mathematical Physics

Chris's sis

@robjohn impressive that work! It's really in my line! I especially enjoyed the first part where I tried to use integrals and I gave up. :-)

skullpatrol

@KevinDriscoll Well, the question has been here for five months without even one comment :-(

Kevin Driscoll

That's true @skullpatrol. I suspect this is because no one knows anything about it. But it could also be it is uninteresting or poorly asked.

skullpatrol

@KevinDriscoll Not even one comment in five months says, to me, move it to MathOverflow :-)

At least you may get a comment...

...you literally have nothing to lose.

:-)

@KevinDriscoll You could start a bounty?

Kevin Driscoll

5:42 AM

@skullpatrol My precious reputation! I have so little.....

but thanks for your advice

I am considering posting to MathOverflow

Right now though I am getting help from a couple of people in person

I don't have time right now to act on outside suggestions, but maybe in thefuture

Chris's sis

5:56 AM

@robjohn btw, your proof above needs no connection to Riemann sums anymore.

robjohn

7:00 AM

I know, it doesn't. It boiled down to the asymptotic
$$
\sum_{k=2}^n\frac1{\log(k)}\sim\frac{n}{\log(n)}\left(1+\frac1{\log(n)}+\frac2{\log(n)^2}+\frac6{\log(n)^3}+\dots\right)
$$

Chris's sis

@robjohn right. Very clever that approach.

@robjohn actually if I remember well, the first limit you computed was given to some university exams (entrance exams).

@robjohn supposing the general term inside the brackets is $k!/\log(n)^k$

robjohn

@Chris'ssis Indeed

skullpatrol

So, if a teacher writes $\frac{1}{0!}$ he's not trying to say never divide by zero :-)

robjohn

7:30 AM

@skullpatrol That would be $\frac10!$

@Jasper: is your blue lighter?

user87637

@robjohn Yes, it is.

skullpatrol

@robjohn Agreed.

robjohn

@Jasper I thought so.

user87637

@robjohn I changed it from steelblue to dodgerblue.

robjohn

@Jasper baseball fan?

user87637

7:34 AM

@robjohn Nope, though I know dodger has something to do with some team.

robjohn

@Jasper It's the Los Angeles team.

user87637

@robjohn Ah, this is a sign that I should apply to UCLA, LOL.

skullpatrol

@robjohn ...one of?

robjohn

@Jasper Indeed. I had thought that that might be the reason for the change.

@skullpatrol The baseball team

skullpatrol

@robjohn yes...

@robjohn Is Anaheim a separate city?

robjohn

7:46 AM

@skullpatrol It is in another county (Orange County)

skullpatrol

@robjohn Thanks, I forgot.

Dominic Michaelis

Does the weierstraß function maps nullsets to nullsets ?

Martin Sleziak

9:03 AM

Should a question be closed, if it is not a duplicate, but one of the answers for the other question answered this question too?

More details here:

in Jury Duty, 4 mins ago, by Martin Sleziak

The question Is product of two closed sets closed? was closed as a duplicate.

Karl Kronenfeld

@MartinSleziak I don't think it's a bad idea, but the link at the top of the duplicate looks funny.

Maybe if one could fine-tune the system so that it links to a specific answer, it would make more sense.

Martin Sleziak

10:23 AM

@KarlKronenfeld If you mean because of the title of the other question, the tile was Product of two sets is closed before I've edited it.

If someone looked only on the title, it seemed more like a duplicate.

what'sup

11:10 AM

hi guys

Karl Kronenfeld

yo

@MartinSleziak That is true, but it is a clash between a very good thing (improving titles) and a questionable thing (closing as duplicate one of a pair of different questions based on provided answers).

Martin Sleziak

@KarlKronenfeld The question is already reopened.

eXtremiity

1:00 PM

Assistance anyone?

Alizter

1:23 PM

@robjohn For some reason I have to wait 10 hours to give you a bounty. I will try not to forget

IvanMatala

hi :)

skullpatrol

hi

IvanMatala

hey skullpatrol hw r u

skullpatrol

f t u?

@IvanMatala Fine thanks. How are you?

what'sup

hi guys , you want me to put a challenge

??

IvanMatala

1:32 PM

im fine too :)

ok heres the challenge

assuming we have this: puu.sh/4eHBW.png

where does the '-k' come from puu.sh/4eHET.png

'negative k'

@skullpatrol do you know explanation where -k comes from???

what'sup

note that in the integral : $$ k \to - k $$

@IvanMatala

Is $$ x(t) $$ is even ??

IvanMatala

um,,, its not given

no information regarding its evenness or oddness

@what'sup you have any idea how did 1 become 2? puu.sh/4eHYG.png

what'sup

Just ask the question or just wait @robjohn @PeterTamaroff

Because i'm busy now

sorry :-)

IvanMatala

sure..

@PeterTamaroff @robjohn

ShuklaSannidhya

1:54 PM

0

Q: How does this shell hit the aircraft?

A fighter aircraft is flying horizontally at an altitude of 1500m with speed of 200m/s. The aircraft passes directly overhead an anti-aircraft gun. The muzzle speed of the gun is 600m/s. We are required to find the angle to the hor at which the gun should fire the shell to hit the...

Is this question more suitable on Math SE?

skullpatrol

@ShuklaSannidhya What book did you find it in?

ShuklaSannidhya

@skullpatrol Why does that matter? [Anyways, it was in my school textbook]...

robjohn

@IvanMatala That is called Fourier Inversion

skullpatrol

@ShuklaSannidhya Which subject?

ShuklaSannidhya

@skullpatrol Physics

robjohn

1:59 PM

@ShuklaSannidhya It looks more like a problem from a math book that one from a physics book. I guess it could be in a beginning physics book talking about acceleration, etc.

skullpatrol

You just answered your own question :-)

robjohn

@ShuklaSannidhya However, most physicists should be able to solve it :-)

skullpatrol

@ShuklaSannidhya Context always matters.

robjohn

I am off to UCLA to proctor an exam. BBL

skullpatrol

later

Alizter

2:17 PM

I've realised that in the unanswered section there are a lot of new users asking questions and then never coming back to them.

skullpatrol

It's called "just give me the answer" >8(

IvanMatala

@skullpatrol hahaha

skullpatrol

:)

user87637

2:53 PM

I have started a new blog, though I won't be writing much.

user87637

Hmm, the kids are all out partying on Friday...

IvanMatala

sure jasper mind sharing it

user87637

@IvanMatala Oh, it's linked from my profile.

IvanMatala

btw, how did a become b? puu.sh/4eKEc.png

user87637

Ah, you should ask an expert. I am only a banana...

Gustavo Bandeira

3:03 PM

@Jasper Hello, Jasper!

user87637

@GustavoBandeira Oh hi.

Gustavo Bandeira

@Jasper How are you?

anon

@IvanMatala linearity; $\int\int A(\tau)B(t,\tau)d\tau dt=\int A(\tau)\int B(t,\tau)dtd\tau$ (if we don't use physics notation of putting dx before the integrand)

ShuklaSannidhya

3:41 PM

Is it possible to find the difference between the x-component of two vector if we have the angle between them and their y-components (The y-component of both the vector are equal to a known constant, but the x-components are unknown).

anon

3:52 PM

make two lines given by the given y-coordinates. for every point on one line, there is another point on the second line that makes the given angle with the first point.

Kevin Driscoll

@skullpatrol Thanks for the bounty on my question.

skullpatrol

@KevinDriscoll No problem pal :)

Kevin Driscoll

@skullpatrol I have actually done quite a lot of things in the months since I asked this question. Is it acceptable that I add more information to my already long question? I'm worried no one will read it if its too long, but they may suggest something that I already know doesn't work if I don't add things.

skullpatrol

@KevinDriscoll Sure, it can't hurt to add more context.

Alizter

4:09 PM

This looks yummy.

skullpatrol

4:38 PM

Hi @TedShifrin how are you?

Ted Shifrin

hi, skull ... doing ok, thanks, and you?

skullpatrol

Fine thanks :-)

Ted Shifrin

Learn anything good today?

skullpatrol

some programming in BASIC

Ted Shifrin

yikes ... who does BASIC anymore? :)

Gustavo Bandeira

4:41 PM

@skullpatrol Why basic?

skullpatrol

because i'm just a banana

:D

Ted Shifrin

Maybe you should be a mango ? :)

skullpatrol

yes, they are sweeter, but they have a pit

Gustavo Bandeira

@skullpatrol A Brad Pitt?

Ted Shifrin

LOL ... so many directions to go ...

skullpatrol

4:44 PM

indeed

Ted Shifrin

I'll see you later, guys. Take care, skull. :)

skullpatrol

later pal

thanks for dropping by

Kevin Driscoll

6:19 PM

Anyone have a copy of Gradshteyn and Ryzhik that I can check against? I think mine has an error in it.

user87637

6:45 PM

@KevinDriscoll Nope, but what book is that?

Kevin Driscoll

Table of Integrals, Series, and Products 7th edition. Its one of the classic table of integrals

so nothing interesting

@jasper

user87637

Well, everything is interesting, lol.

Alizter

Except essays about statistics

Kevin Driscoll

@Alizter WHAT!? Theres a famous Econ paper called 'A Market for Lemons' that is essentially an essay about statistics and its quite good

user87637

Well, statistics can be interesting too, lol.

Alizter

6:47 PM

Writing a statistics paper about asking people what their favorite angle is is not exciting.

I like the mathematics in statistics.

But the social elements I dislike very much.

Why do people like samples better than censi? Ermm preference?

user87637

Statistics at the higher levels can be very mathematical.

user87637

But at the lower levels, it is quite boring.

user87637

@Alizter Never even saw censi before, lol.

Alizter

@Jasper Censi sounds cooler than Censuses

user87637

@Alizter Sounds like sensei, lol.

Alizter

6:51 PM

@Jasper Oh dear is that something rude?

user87637

@Alizter No, it means teacher in Japanese I think...

user87637

@Alizter Is it still true that you can't apply to both Cam and Ox in the same year?

Alizter

@Jasper No idea i can't apply for another 2 years anyway

user87637

@Alizter I have a feeling that is true, hmmm... Anyway if you can't get in Cam or Ox, go for Imperial and Warwick like I said...

Alizter

Imperial is difficult because of being in london

but im taking note of warwick

Kevin Driscoll

6:54 PM

@ALizter are you in the UK?

Alizter

@KevinDriscoll I'm checking... YEP. Google says I am so that must be true!

Kevin Driscoll

@Alizter Haha, if Google says so....

user87637

@Alizter Google said I was in the US though I was not, lol.

user87637

Google is not always correct you know, lol.

Alizter

Chinese Google says that there exists no other countries.

Kevin Driscoll

6:55 PM

@Alizter Do Ox-bridge use the same general system as the other universities where you apply and they give you a 'offer' in terms of the number of A-levels you get etc etc?

user87637

@KevinDriscoll Yes. You usually need the STEP papers too.

Alizter

@KevinDriscoll The only difference between them and other universities is entry requirements. They are slightly more strict on who can come in

user87637

The STEP papers are very challenging and have an olympiad flavour.

user87637

You can practise past year papers I think.

Alizter

@Jasper All they need is A*AA in A-level.

Kevin Driscoll

6:57 PM