Mathematics - 2012-11-20 (page 1 of 3)

Argon

12:00 AM

It has interesting residues to figure out! :)

jdoe

I can't see how though

Argon

@jdoe Wait, maybe not - I notice that the lower bound is 1

jdoe

could it be holomorphic on the whole complex plane with lower bound 1?

$ = \int_1^\infty e^{(x-1)(\log t) - t} dt$

I suppose we could argue that whatever complex constant x is, -t dominates (x-1)(log t) so basically it's e^{-t} for large t

Argon

@jdoe Perhaps using $(2)$ from mathworld.wolfram.com/IncompleteGammaFunction.html

jdoe

$$\frac{e^{(z-1)\log t}}{e^t}$$

Argon

12:10 AM

Your integral with the lower bound of $1$ can be written as $\Gamma(x, 1)$ if that is helpful (Incomplete Gamma Function).

jdoe

well I think we can prove it directly because if t = e^y then that fraction (the integrand) is <= 1 whenever e^y/y > |z-1|

oh wait that's not enough because the limit is infinite

I mean the top is "polynomial" and the bottom is exponential so it must converge

Argon

No poles in the integrand...

jdoe

oh yeah the integrand is holomorphic so its primitive will be

Argon

Also, while this is not rigourous, it is notable that for every $t$ between $1$ and $\infty$, the function is never indeterminate like in the regular gamma function w/ a lower bound of $0$.

jdoe

what you mean like intederminate?

Argon

12:15 AM

i.e. $\frac{e^0}{0}$ in the integrand when $\Gamma(0)$

jdoe

oh

so the integral definition of gamma works for all complex z with Re z > 1?

Argon

I believe so. 1 sec

@WillHunting Hi again. How was breakfast?

user19161

@Argon OK. I might go out to the park later for a walk. I have not been to that one for over a year...

J. M.

@jdoe $\Re z > 0$ if memory serves...

Argon

@WillHunting Go ahead!

jdoe

12:17 AM

@J.M., ah, that's good

@J.M., trying to continue it with this sum + integral is actually veryhard

because I cant show the integral converges

it obviously does but I don't know how ot proveit

user19161

@jdoe What did you use to plot that?

Argon

@WillHunting Cf. en.wikipedia.org/wiki/File:GammaAbsSmallPlot.png

jdoe

@WillHunting, I got it from wikipedia

> English: A 3D plot of the absolute value of the Γ function, self-made in Mathematica (prior versions in MuPAD)

user19161

@jdoe HAHAHAHAHA

jdoe

what

user19161

12:21 AM

Nothing, I just did not expect it. =)

jdoe

Let s be any complex number show that $$\int_1^\infty e^{-t} t^s dt$$ converges

Argon

Hmmm... I need to get smarter at this.

jdoe

ah i got it

let s = x + iy

Argon

Ok

user19161

@j.m. I am wondering if you have any math books that are out of print today.

jdoe

12:24 AM

$|t^s| = |t^x||\exp(i y \log(t))| = |t|^x$

user19161

@amwhy I got the Strunk and White badge now! That's for editing 80 posts.

jdoe

so the integrand $\le |e^{-t}||t|^{\Re{s}}$

Argon

@jdoe $\log z= \operatorname{Log} |z|+i\theta$

@WillHunting Congratz

jdoe

I showed the integrand is bounded by $x e^{x \log(x)} e^{\log u - u}$ with substitution $xu = t$

So just need to show that $$\int_1^\infty e^{\log u - u} du$$ converges

J. M.

@WillHunting Yes, mostly yellowed...

One of these days, I should scan them before they become dust...

jdoe

12:31 AM

could split that into $\frac{u}{e^{u/2}}\frac{1}{e^{u/2}}$ where the first factor is < 1 for large enough u and the second factor decreases exponentially which we need for the integral to converge

user19161

@J.M. Could you name one famous one?

J. M.

I think Jahnke-Emde is pretty famous...

jdoe

wait a sec how do we know the gamma integral converges for anything..

Let $$\Gamma(s) = \int_0^\infty e^{-t} t^{s-1} dt$$ this integral converges for $\Re (s) > 0$.

amWhy

@WillHunting Yes, I've got that badge. I think I may even have a gold badge for editing 500 some posts awhile back!

Argon

@jdoe $\Gamma(1)$ converges trivially.

We show the rest of the integers converge by repeated integration by parts

user19161

12:36 AM

@amWhy Yes, I just saw that you did over 700!

Argon

@jdoe $\Gamma(1)$ converges trivially.

jdoe

@Argon, but what about between integers and complex values?

Argon

We show the rest of the integers converge by repeated integration by parts

amWhy

@WillHunting Wow, didn't realize it was that high, and I was gone for nearly a year (from math.se).

user19161

@amWhy You can check this out on the users page, just click on the editors tab and sort by all time.

amWhy

12:37 AM

@WillHunting I learned LaTeX by editing

5

user19161

@amWhy Haha. Do you install TeX yourself on your own machine?

Argon

@jdoe The paper I linked you to yesterday had a proof

@amWhy Me too :)

amWhy

@WillHunting I've got it installed, but can't figure out how to work with it. I've got to get back on that task!

jdoe

@Argon, it says:

> Clearly, the second integral converges for any complex z

lol :D

Argon

@jdoe Hahaha

user19161

12:39 AM

@amWhy Haha, OK you can ask me if you run into problems. I recommend TeX Live over MikTeX.

jdoe

at least it's clear to someone

Argon

@jdoe "It can be shown..."

@WillHunting LyX works for me but MikTeX never does - though it is needed by LyX!

amWhy

@WillHunting Thanks! I'll look into TeX Live. I've got MikTeX installed, but haven't really tackled it yet. I need to learn how to "begin" a document (headers, etc.). I know the math part, it's the rest that I need to learn.

user19161

@Argon Oh, LyX is an editor and MikTeX is a TeX distribution, so they are different things. I use TeXworks and TeX Live.

Argon

@WillHunting LyX forced me to install it

user19161

12:42 AM

@Argon To install MikTeX?

Argon

@WillHunting Yep

jdoe

how do i show there's a t' such that for all t >= t', $k \log t < t$

user19161

@Argon Anyway, I don't like LyX like I told you.

Argon

@WillHunting I know. But I do!

user19161

@Argon Good for you. I use TeXworks like the great anon!

Argon

12:43 AM

@WillHunting The great anon!

user19161

@Argon TeXworks: it just works!

user19161

HAHAHAHAHA

Argon

@WillHunting You sound like you work for them :)

user19161

@Argon I did try LyX once, it was just too weird, neither here nor there but in between.

Argon

@WillHunting So? It's an editor that allows TeX to be inserted as necessary!

user19161

12:45 AM

@Argon It's more like between a word processor and an editor.

Argon

@WillHunting And awesome

user19161

@Argon I also love the user interface of TeXworks, code on the left half and output on the right, and the ability to jump to and fro.

Argon

@WillHunting Never actually used it, so I cannot comment

user19161

@Argon The great Jonas uses Emacs.

Argon

@WillHunting Yay

@WillHunting What is it?

J. M.

12:47 AM

Hi Amy!

user19161

@Argon An editor that is even harder than TeXworks.

user19161

@J.M. Hehe, at first I was wondering who you were referring to, then I learnt about it a few days ago...

Argon

@WillHunting When you are that good, I guess you can use it!

user19161

@Argon When you told me your last name, I understood why you are called Argon. =)

Argon

@WillHunting Marilia made the same realization :)

user19161

12:57 AM

@Argon I should be called Joy then. =)

Argon

@WillHunting Hahahaha Ya

jdoe

I'd like to study the pochhammer contour continuationof beta

user19161

By the way, what happened to Mr Mick?

Argon

@jdoe ...I would too... If I could understand it!

That's a problem with math - I don't understand it :)

@WillHunting A very good question!

jdoe

@Argon, yes wwe can both understand it should be easy

$$
\displaystyle (1-e^{2\pi i\alpha})(1-e^{2\pi i\beta})B(\alpha,\beta) =\int_C t^{\alpha-1}(1-t)^{\beta-1} \, dt. $$

C is anything like that

bye

Argon

1:04 AM

Bye

robjohn

1:18 AM

@J.M.: Hey there... I've been away from chat for most of the day.

Are things slow here for others, or is it just local?

Link

Local

robjohn

@Link drat. I end up waiting for stackexchange.com and mathjax.org, so I thought it might be something on the other end.

Argon

@robjohn Do you have any tricks for deciding on what contour to use in integration?

J. M.

@robjohn Hey. Nothing unusual, it looks.

robjohn

@Argon It is a matter of looking at singularities, branch cuts and where the function is well behaved.

@Argon and sometimes, you can use different contours to get the same integral.

Argon

1:23 AM

@robjohn I always end up choosing a bad contour and then I don't know if it's my fault or the function on the contour :)

J. M.

(i.e., don't have your contours cross regions where the function is wonky...)

robjohn

@J.M. important safety tip :-)

Argon

@J.M. I learned that the hard with with hyperbolic functions, hahaha!

They act like circular trig functions on the complex axis!

So I often use a big rectangle for those

J. M.

@Argon That's actually typical for a number of special functions: oscillatory in one region, exponentially growing/decaying in another...

(More often than not, you want the regions where it's decaying)

Argon

I have trouble w/ branch cuts and points as well...

How do you know where the cut is? e.g. with $\log$, must the cut run along the real axis?

J. M.

1:28 AM

@Argon The principal branch of the logarithm is cut on the negative real axis.

Argon

@J.M. Why is that? Could I integrate it so I avoid, say, the positive complex axis instead?

J. M.

@Argon It's more a convenience than anything. Of course, there's nothing preventing you from spinning the branch cut around so it ends up along a different ray...

Argon

@J.M. So should I think of it as a spiral, like the Riemann Surface, where I just need to make the function 1 to 1 by cutting somewhere?

J. M.

Yes, the Riemann surface is an excellent way of looking at it. Since the logarithm is multibranched, you need to slice a particular section to get the one-to-one feature.

Argon

Neat. Because $a^b = \exp (b\log a)$, it follows that some exponential functions are multivalued as well. Which ones are? Why don't I take branch cuts for $e^x$?

J. M.

1:33 AM

So, you'll want something like $\log |z|+i(\arg(z\exp(i\theta))-\theta)$ if you need the cut to be somewhere else...

@Argon The exponential function is entire/analytic/holomorphic, so no cuts...

@Argon Yes, power functions for noninteger powers should also inherit the logarithm's branch cut.

Argon

@J.M. So is $\exp$ the exception?

J. M.

@Argon I'm making a distinction between exponential functions and power functions, to be clear. For $a^b$, $a$ is the one varying in a power function. For an exponential function, it's $b$ who's varying.

Argon

@J.M. I see, thanks. So $f(x)=x^a$ has cuts?

J. M.

@Argon For noninteger $a$, yes.

Argon

@J.M. Why not integers?

user19161

1:39 AM

@J.M. Wow, this is the first time in my life I see "noninteger" I think! =)

J. M.

@Argon If they weren't, polynomials wouldn't be entire...

A constant function $z^0$ is obviously entire; $z$ can be shown to be entire; and the product of two entire functions is also entire, so...

user19161

Interestingly, Lang uses the term "entire ring".

Argon

@J.M. That is very strange! :)

J. M.

For, say, $1/z$, it's meromorphic (pole at zero), but no cuts...

Argon

But $z^1 = \exp(\log |z|+i\theta)$! Why does the argument not make multiple values?

I can say I'm still not entirely comfortable manipulating complex numbers like I am with reals...

user19161

1:43 AM

@argon Check your mail.

Argon

@WillHunting Recieved, sir!

user19161

@Argon receive

Argon

@WillHunting I wish I understood something there :'(

user19161

@Argon It is trivial, just thought I would send you for fun. You can read it in future. =)))

user19161

@Argon The only complex analysis I remember now is the definition of a complex number. =)

Argon

1:46 AM

@WillHunting $z=x+iy=Re^{i\theta}$. QED

user19161

@Argon Hahaha.

J. M.

@Argon ...and what's $|z|\exp(i\arg\,z)$ supposed to be? :)

(remember the exponential function's periodicity as well)

user19161

@argon I can't view what you sent me as I don't have an account, you can send me the PDF if you want.

J. M.

On that note: you should be able to recall one rather inconvenient feature of the polar coordinate system...

user19161

@J.M. Which is?

J. M.

1:50 AM

@WillHunting To illustrate: why are $(1,\pi/4)$ and $(1,-7\pi/4)$ the same point?

Argon

@J.M. Hmmm... what even is $\exp(i\arg z)$? $z$ where $|z|=1$? (Don't laugh too hard)

user19161

@J.M. OIC. Yes I can still understand that. =)

J. M.

Okay, answer my question on polar coordinates first. :) (Seems Will got it.)

Argon

Principle valueness

Periodicity

J. M.

$\exp$ is $2\pi i$-periodic, yes?

Argon

1:52 AM

Yep

@WillHunting Check your electronic inbox!

J. M.

So, whatever branch of the phase you pick, it shouldn't matter within the exponential.

Thus, you always have the polar decomposition $z=|z|\exp(i\arg\,z)$

Argon

So let us now consider $\sqrt{z} = \exp\left(\frac{1}{2}(\log |z|+i\arg z)\right)$

user19161

@Argon Received. Note the spelling again!

Argon

@WillHunting Spelling??

$\sqrt{|z|}\exp(i\arg z/2)$?

J. M.

@Argon You lose the periodicity now; think of what happens after adding or subtracting an integer multiple of $2\pi$ from the argument...

user19161

1:55 AM

@Argon The spelling of "receive" which is not "recieve".

Argon

@WillHunting Meh

@J.M. $\exp(i\arg z/2+i\pi)$?

J. M.

@Argon That'd correspond to the negative square root, no?

Argon

@J.M. Yep :)

Cool! $\exp(i\pi)=-1$

Thanks!

J. M.

(It should now be clear why we have to pick which of the positive or negative square root should be principal for our purposes...)

Argon

Cool, thanks :)

My favourite show, Midsomer Murders, is on now. Good night!

J. M.

2:00 AM

Anyway, I must go. See y'all later.

Argon

@J.M. Bye sir!

user19161

@Argon Oh noes, now you call everyone sir like iyengar!

Argon

@WillHunting Iyengar?

user19161

@Argon OK go watch your TV.

user19161

@Argon A user who came to chat long ago.

Argon

2:01 AM

@WillHunting Hahaha :)

@WillHunting He said sir then, I presume

user19161

@Argon The logic is obvious. QED.

Argon

@WillHunting Enjoy Weierstrass substitution. $\blacksquare$

Good night

user19161

@Argon See you in your dreams.

leo

2:17 AM

Are manifolds normal topological spaces?

user19161

@leo Yes, of course.

user19161

The usual definition is that it is a topological space that is Hausdorff and locally homeomorphic to Euclidean space.

Link

Hello again

user19161

@leo Oh, sorry I misread the normal there. I thought you meant ordinary.

wj32

@leo: do you require manifolds to be second countable?

user19161

2:27 AM

@wj32 Ah, I missed that out too. =)

user19161

By the way @wj32 what does your username mean?

wj32

nothing

why does it have to mean something

Gustavo Bandeira

3:30 AM

0

Q: What's the meaning of the exponent here?

I'm reading A First Course in Logic: An Introduction to Model Theory, Proof Theory, Computability, and Complexity. The graph of $f: A \rightarrow B$ is the subset of $A × B$ consisting of all ordered pairs $(a, b)$ with $f (a) = b$. If $A$ happens to be $B^n$ for some $n ∈ N$, then we say...

user19161

4:16 AM

@GustavoBandeira Ah I wanted to answer that one but it's taken up already. =)

John Smith

if I know the last M digits of fibonacci(N) is there a way I can extract Fibonacci(N)%b from this for arbitrary b?

wj32

4:41 AM

@WillHunting: you've studied algebraic topology, right?

N3buchadnezzar

7:28 AM

@robjohn hi!

robjohn

@N3buchadnezzar Hey there!

Matthew Wong

does anyone here know how to find the upper and lower bounds of a polynomial? (precalculus math)

robjohn

what's up?

@MatthewWong for an odd order polynomial, there won't be either

N3buchadnezzar

@robjohn I have an integral, that I am supposed to optimize using linear algebra. Got time give it a glimpse? My book does not cover it

robjohn

@N3buchadnezzar what integral?

Matthew Wong

7:30 AM

@robjohn by odd do you mean number of coefficients or the power?

robjohn

@MatthewWong the order of a polynomial is the degree of the highest term: $x^3+x$ is third degree

Matthew Wong

?! if there isn't either then why does the textbook list it as a question?

robjohn

@MatthewWong The upper and lower bounds for the real roots of the polynomials

Matthew Wong

oh i guess i should've mentioned that too

my bad :P

robjohn

@MatthewWong what does Example 2 say?

N3buchadnezzar

7:35 AM

$$ \int_0^1 \left| t^4 - a - bt \right|^2 \,\mathrm{d}x$$
Find $a$ and $b$ to minimize.

Matthew Wong

i wish i knew or else i wouldn't have to ask. (i tried doing a google search of the textbook and the page wasn't listed in the preview)

this is a photocopy made by the teacher

N3buchadnezzar

According to the solution to a few similar problems they use it as a inner product, to calculate the projection from the space a + bt onto the function t^4. But I a having problems figuring out the details, as the integral does not define any inner product?

robjohn

@N3buchadnezzar can you compute $\frac{\partial}{\partial a}$ and $\frac{\partial}{\partial b}$ of that integral?

@MatthewWong what is the text book?

Matthew Wong

@robjohn Precalculus: With Unit Circle Trigonometry

N3buchadnezzar

@robjohn No?

robjohn

7:45 AM

$\frac{\partial}{\partial a}\int_0^1(t^4-a-bt)^2\,\mathrm{d}t=-2\int_0^1(t^4-a-bt)\,\mathrm{d}t$ and $\frac{\partial}{\partial b}\int_0^1(t^4-a-bt)^2\,\mathrm{d}t=-2\int_0^1(t^4-a-bt)t\,\mathrm{d}t$

set both to $0$ and solve for $a$ and $b$

N3buchadnezzar

I thought one needed the abs value

robjohn

@N3buchadnezzar what is $|x|^2$?

Matthew Wong

@robjohn nevermind i found a powerpoint that helps me

robjohn

@MatthewWong yeah, without knowing what Example 2 says, it is hard to know what to do.

N3buchadnezzar

@robjohn =) Gives me $a=-4/5$, $b = 1/5$ and $I_{a,b} = 4/225$

I guess I have to ask my teacher about the linear algebra approach to this problem

Matthew Wong

7:50 AM

fuck me...i have to list all the factors of 100 and synthetically divide them...

robjohn

@N3buchadnezzar well, it is the inner product of $t^4-a-bt$ with itself

@MatthewWong to bound the real roots?

@MatthewWong that sounds like what you'd do to find the rational roots

Matthew Wong

yeah except it doesn't matter if theres a remainder or not

just quoting here

"Divide f (x) by x - b (where b > 0) using synthetic division. If the last row containing the quotient and remainder has no negative numbers, then b is an upper bound for the real roots of f (x) = 0."

"Divide f (x) by x - a (where a < 0) using synthetic division. If the last row containing the quotient and remainder has numbers that alternate in sign (zero entries count as positive or negative), then a is a lower bound for the real roots of f (x) = 0.
"

sounds a bit too easy tough o_o

*though

Nimza

8:43 AM

Good day!

Hi @robjohn! Did you hear about nonnegative distributions (generalized functions)?

robjohn

@Nimza what about them? did something happen in the news?

Nimza

@robjohn no, I can't find anything on them. But I was told that there's some Bochner characterisation theorem for them...

robjohn

@Nimza where are you seeing them?

Nimza

@robjohn yesterday on seminar orator was speaking about them, but he thought that such basic things are classical and ommited them

robjohn

@Nimza I don't know if this has anything to do with it.

Nimza

8:56 AM

@robjohn yeah, looks like very similar to it! And I've found something in Vladimirov's book "Generalized functions in mathematical physics"

robjohn

@Nimza Those distributions are a bit different, but I guess in certain usages, they could be used in a similar way

Nimza

@robjohn aha, thank you for help :)

N3buchadnezzar

9:26 AM

Hmm

What does $\tilde{x}(t)$ mean ?

robjohn

10:01 AM

@N3buchadnezzar It depends on the context

N3buchadnezzar

10:54 AM

Kay

robjohn

@N3buchadnezzar what is the context that you have?

N3buchadnezzar

IT is in regard to functional analysis, and metric spaces.

Problem 3

math.ntnu.no/~stacey/documents/tma4145/tma4145mt05ne.pdf

robjohn

11:25 AM

@N3buchadnezzar In that question, it just seems to denote a second value. That is $x$ and $\bar{x}$ are like $x_1$ and $x_2$ or $x$ and $x'$.

@N3buchadnezzar admittedly, that was not one of the uses I've seen for $\bar{x}$

Wow, we are down to 4...

Alexei Averchenko

12:36 PM

is there a picture of Michael Spivak out there?

user1230219

1:21 PM

Hi, I'm currently a student in an online AP BC calc class, and I'm having a little trouble with a question about derivatives and parametric equations - I emailed by teacher, but I think she's on an unofficial early thanksgiving break -.- (We are assigned enough work regardless to have to work until friday, but she obviously won't be there - and all assignments are due then)

Anyways, the problem is written as follows:
"The position of an object is described by the parametric equations x=ln t and y=5t^2. What is the acceleration of the object in /sec^2 when t=2?

I've been wondering the whole time what I should be calculating - my sister says I should be calculating d^2y/dx^2, but personally I think I should be calculating sqrt(d^2y/dt^2+d^2x/dt^2) for t=2 - I'm a little confused, because I thought acceleration had direction as well, meaning that a single scalar would not accurately describe the acceleration of a particle with motion described in x and y axis?

Thanks guys.

jdoe

1:36 PM

hello

user1230219

Hi

jdoe

> The position of an object is described by the parametric equations x=ln t and y=5t^2. What is the acceleration of the object in /sec^2 when t=2

hmm

is acceleration a vector (with x,y components) or a magnitude?

> because I thought acceleration had direction as well, meaning that a single scalar would not accurately describe the acceleration

I agree with you here

let's say acceleration is a vector with x and y components

@user1230219, the position of the object at time t is (ln t, 5t^2) so it's velocity is ( d/dt ln t, d/dt 5t^2 )

user1230219

And acceleration is the second derivative correct?
The question is, is my teacher really looking for this, or simply looking for d^2y/dx^2?

I mean, the magnitude of acceleration.

calculating the second derivative for the x and y components, I get
d^2x/dt^2=e^x
d^y/dy^2=10

jdoe

do you agree velocity is ( d/dt ln t, d/dt 5t^2 )

user1230219

To be quite honest, I'm not sure - intuitively yes, I guess. Are we missing a direction component here?

jdoe

1:49 PM

ok so lets start at basics

We have an object whose position at time t is given by (x(t),y(t)) and those are the x,y coordinates

so in this case the object is at (ln 2, 5*4) when t=2

user1230219

Okay.

jdoe

and at t=3 it's at (ln 3, 5*9)

so in one second it has moved this much: (ln 3 - ln 2, 5*(9-4))

so for t between 2 and 3, it's velocity is about 0.4 m/s in the x axis and 25 m/s in the y-axis

user1230219

I see

jdoe

ok good!

so do you know why I just said velocity is 'about'

rather than exactly

user1230219

It's not instantaneous?

jdoe

1:56 PM

yeah

N3buchadnezzar

$f(t) \cdot \overline{f(t)} = ?$

peoplepower

complex conjugate?

jdoe

@user1230219, so what would give the exact velocity at some time t?

N3buchadnezzar

@peoplepower Indeed, is there any easier way to write it ?

$$|f(t)|^2$$

peoplepower

@N3buchadnezzar Is it not $|f(t)|^2$?

N3buchadnezzar

1:58 PM

Silly me

peoplepower

Silly me too, I forgot to square it.

user1230219

@jdoe, sorry for my delayed reponses - I made a post in math.stack exchange and am reading feedback

the exact velocity at some time t would be (x'(t),y'(x)), correct?

jdoe

@user1230219, no y is a function of t

user1230219

Er, typo there

N3buchadnezzar

@peoplepower Thanks anyway =)

user1230219

2:05 PM

Correction:(x'(t),y'(t))

and the acceleration would be (x''(t),y''(t))?

jdoe

@user1230219, yeah differentiating with respect to time, it's velocity being how the position (in each coordinate) changes with time

@user1230219, so you can compute it ( d/dt ln t, d/dt 5t^2 ) = (1/t, 10t)

user1230219

The only problem is that I am requested to give units in the form m/sec^2 - again, I'm guessing that they want magitude correct?
http://math.stackexchange.com/questions/241312/acceleration-of-a-particle-described-by-parametric-equations

jdoe

so what is the acceleration

user1230219

The second derivative?

jdoe

yes but compute it

user1230219

2:09 PM

well, I have two parametric equations
x=ln t and y=5t^2
d^2x/dt^2=-1/(t^2)
d^2y/dt^2=10

jdoe

great so acceleration is?

user1230219

so i evaluate for t=t, the eppression sqrt(10^1+(-1/4)^2)?

jdoe

no

user1230219

Well, the problem is phrased as wanting units in m/sec^2...

jdoe

position is (ln t, 5t^2), velocity is (1/t, 10t)

user1230219

2:12 PM

Hey, I'm really sorry - the bell just rang, so I have to get to my next class
thanks for helping :)

jdoe

2:44 PM

darn I forgot t look up the pochhammer thing at the library today

hi

N3buchadnezzar

And then Thor used the POCHHAMMER! And peace once lay over Valhalla.

anon

3:30 PM

Artin defines rings with unity right?

Jayesh Badwaik

@N3buchadnezzar and loki is now spreading the word?

N3buchadnezzar

@JayeshBadwaik y

jdoe

if f is in L^1 then is sup_[a,b] |f(t)| defined?

anon

sup |f| is the L^1 norm right?

jdoe

I don't even know :(

anon

3:44 PM

yes I believe so

wait, that's the L^infty norm

the L^1 norm is int |f| dt

jdoe

so ill have to say f is in L^1 and bounded

but how would I show its fourier transform $$\hat f (\lambda) = \int_{-\infty}^\infty f(t) \exp(- i \lambda t) dt $$ is bounded

oh that's easy just use |integral fg| <= integral |f||g|

user19161

4:08 PM

@anon Yes.

user19161

@N3buchadnezzar Haha, Andrew is your lecturer?

Jayesh Badwaik

@WillHunting Hi. Why did you ditch Ubuntu? (just asking, because I just upgraded my dad's computer to Ubuntu 12.04) Its pretty good. I even donated some money to Canonical as a fee.

jdoe

Can you help me show $\hat {\hat f(t)} = 2 \pi f(-t)$

using $$E(x) = e^{-\frac{x^2}{2}}$$

I think it just comes from $\hat E = \sqrt{2 \pi} E$ and $\int \hat f g = \int f \hat g$

Jayesh Badwaik

Parsevals

jdoe

4:24 PM

I need $$\hat {\hat f(t)} = 2 \pi f(-t)$$ to prove parseval

Jayesh Badwaik

one way is to use the fact about orthogonality of $exp(-i\lambda_1 t)$ and $exp(-i \lambda_2 t)$ , $\lambda_1 \neq \lambda_2$ and then some computations.

N3buchadnezzar

@WillHunting He is on vacation, but if he werent he would be.

jdoe

so far I got $$\int \hat f(t) E(t/R) = R \sqrt{2 \pi} \int f(\lambda) \hat E(\lambda/R)$$

as R tends to infinity LHS tends to integral of f hat, but I don't know what the RHS does

oops shouldn't have a hat on the RHS removing it is what gave me the 2pi

should I replace f with something orthogonal times f?

this jst gets more wrong the more I look at it

user19161

4:46 PM

@JayeshBadwaik (1) The default desktop, Unity, is unresponsive and ugly, even though one can install GNOME on top. (2) The only themes available in a default install are ugly, and others don't work well with Unity. (3) It cannot display the graphics driver correctly when other distros can. (4) All its innovations, like overlay scrollbars and HUD, are not to my liking. (5) I like independent distros more than dependent ones.

Jayesh Badwaik

@WillHunting Hmm. Well, my dad uses KDE, so :D

user19161

@anon So does Bourbaki, Cohn, Rowen, Lang, Jacobson, MacLane. But not Herstein, Fraleigh, Grillet, Knapp, Malik.

user19161

@JayeshBadwaik Then just install Kubuntu instead of Ubuntu.

Jayesh Badwaik

@WillHunting Its the same thing, except for installers, also, I had already installed the stuff, just upgraded in situ. I dislike clean installs.

@jdoe hmm, you are tying yourself up in tangles, write the rhs as a fourier transform and then again some similar stuff and you should be done.

user19161

@JayeshBadwaik Really? I like clean installs.

jdoe

4:51 PM

i can't do it

Jayesh Badwaik

@WillHunting Dude, the first rule they taught me at RHEL training course (which I attended for fun) was that you should avoid re-installs as far as possible. (Not giving any authoritarian credence to it. But this practise is actually true, you don't lose settings, you don't lose stuff, you basically hit the ground running.

jdoe

why avoid reinstalls

user19161

@JayeshBadwaik The first thing I want to say about that is, it's all a matter of preference.

Jayesh Badwaik

@jdoe check my updated answer.

jdoe

where

Jayesh Badwaik

4:54 PM

1 min ago, by Jayesh Badwaik

@WillHunting Dude, the first rule they taught me at RHEL training course (which I attended for fun) was that you should avoid re-installs as far as possible. (Not giving any authoritarian credence to it. But this practise is actually true, you don't lose settings, you don't lose stuff, you basically hit the ground running.

user19161

@JayeshBadwaik Upgrade if you don't want to reconfigure your system. Reinstall if you want to reconfigure your system. QED.

Jayesh Badwaik

@WillHunting And you should not have to reconfigure your system too often.

user19161

@JayeshBadwaik Practice is the noun and practise is the verb in BrE.

Jayesh Badwaik

@WillHunting hmm.... too late to modify now. Next time. Thanks.

user19161

@JayeshBadwaik I like to reconfigure once every two years when a new Debian release is out. Upgrades can go wrong, and you may need to download even more stuff and it may take even longer than a fresh install.

user19161

4:57 PM

If you are in the middle of an upgrade and the internet connectivity is not good, you are done for.

Jayesh Badwaik

@WillHunting Depends on the system. Are you aware of the / -> /usr shift?

user19161

Maybe ten hours later when you are done downloading, the upgrade fails and you cry.

user19161

@JayeshBadwaik I know Fedora is changing its filesystem.

Jayesh Badwaik

@WillHunting Well, thing is in ArchLinux, I was able to do it without reinstallation (without anything special actually), which shows how good upgrades can be when done well.

Also, I never had problems upgrading Debian or Ubuntu.

I have upgraded seamlessly from 8.04 to 10.04 to 12.04

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$Mathematics$ Mathematics