Post by: vyper on August 28, 2002, 03:44:55 am
Here I was downloading files on Kazaa last night, blissfully unaware of anything wrong. I download a flile claiming to be a Playboy... image. I go to the "My Shared Folder" directory only to find a file with a name nothing to do with what it was meant to be. (Filename: "lesbia orgy britney spears limpbizkit... korn...." It keeps on going) :mad2: Not only that, can't delete it, I can't rename it, I can't even get properties on it. All I can do is move it about the place.
I even gave in and tried to open it but all programs said the file didn't exist.
:( HEEELP!
Post by: Reaper on August 28, 2002, 04:00:47 am
Post by: Nico on August 28, 2002, 04:06:16 am
Quote
Originally posted by Reaper
Try dos... and thread will probably be locked, cos u mentioned kazaa...
nothing's wrong with Kazaa until you talk about d/ling Warez. Anyway, I had the same pb once, with a fake MP3 I think, deleting under DOS did the job indeed.
Post by: vyper on August 28, 2002, 04:06:33 am
Post by: Shrike on August 28, 2002, 04:07:14 am
:o :o :o :o
Post by: Reaper on August 28, 2002, 04:08:16 am
Post by: heretic on August 28, 2002, 04:08:55 am
First, don't download anythign small from KazAA.
Second, run EVERYTHING through a virus scanner
third... you're a pr0n freakboy!!!
:lol:
Post by: Reaper on August 28, 2002, 04:09:15 am
Quote
Originally posted by Shrike
God, this would have to be the most embarassing thing ever....
:o :o :o :o
Yah... he was probably DLing pr0n... why do u think that kind of file appeared:D :D :D
now lets help the d00d... and when he fix it... lets make fun of him:lol:
Post by: Kamikaze on August 28, 2002, 04:09:18 am
Post by: Shrike on August 28, 2002, 04:09:41 am
Post by: Shrike on August 28, 2002, 04:10:13 am
Quote
Originally posted by ReaperWell duh, he pretty much says as much in the first post.
Yah... he was probably DLing pr0n... why do u think that kind of file appeared:D :D :D
Post by: Kamikaze on August 28, 2002, 04:11:34 am
Post by: Reaper on August 28, 2002, 04:11:56 am
Quote
Originally posted by Shrike
Well duh, he pretty much says as much in the first post.
Playboy ain't pr0n... it's erotic... Theres a difrance
Post by: Kamikaze on August 28, 2002, 04:12:39 am
Quote
Originally posted by Shrike
Lying in this situation is perfectly acceptable. :lol:
desperation can do lots of things.... ;)
Post by: Reaper on August 28, 2002, 04:12:55 am
Post by: vyper on August 28, 2002, 04:16:17 am
Post by: Shrike on August 28, 2002, 04:18:24 am
Quote
Originally posted by Reaper"lesbia orgy"?
Playboy ain't pr0n... it's erotic... Theres a difrance
Sounds like Porn to me.... :p
Sheesh, if you want porn that badly, go to www.google.com
;)
Post by: Reaper on August 28, 2002, 04:21:14 am
Quote
Originally posted by Shrike
"lesbia orgy"?
Sounds like Porn to me.... :p
Sheesh, if you want porn that badly, go to www.google.com
;)
Or you could try [censored]
There's a nice free pr0n on the site :lol: :lol: :lol:
Post by: vyper on August 28, 2002, 04:32:11 am
Post by: heretic on August 28, 2002, 04:55:46 am
Post by: Nico on August 28, 2002, 05:12:44 am
Of course nobody else who posted on this thread never did, right? of course not :rolleyes: I did, and sometimes I still do :devil: Rah!!!! shocking!!!!!
Post by: Sandwich on August 28, 2002, 05:15:02 am
Vyper, firstly, what OS are you running?
If 98 (ME might have it, too), Start > Run... > msconfig {enter}. In the program that pops up, go to the last tab, on the right - I think it's "Controlled Startup" or something like that. Look through the list there, and uncheck anything that says kazaa or "speed up". "Speed up" is a proggie that comes with the latest version of Kazaa Lite (probably with Kazaa normal, too). Anyways, you don't want either of them running on startup for now.
If you're running 2000, I'll walk you through it on ICQ - what's your UIN?
Secondly, change the columns in Kazaa so that the Title column is at the end - it means literrally nothing, and is the reason why you tried one file and got a different one. Put "Filename" first - it's usually an accurate representation of what the file truly is.
The reason you couldn't move or delete that file is probably because it was "in use" by Kazaa, which doing a clean reboot will solve. That is, assuming that it isn't a virus... :-/
Post by: Shrike on August 28, 2002, 05:20:12 am
Quote
Originally posted by venom2506Bah, the entire situation has comedic value.
He downloaded pr0n, big deal :rolleyes:
Of course nobody else who posted on this thread never did, right? of course not :rolleyes: I did, and sometimes I still do :devil: Rah!!!! shocking!!!!!
The funny thing is, just as I was about to reply again, my computer locked up. :lol:
Post by: Reaper on August 28, 2002, 05:23:32 am
Quote
Originally posted by venom2506
He downloaded pr0n, big deal :rolleyes:
Of course nobody else who posted on this thread never did, right? of course not :rolleyes: I did, and sometimes I still do :devil: Rah!!!! shocking!!!!!
I never said i didn't
Post by: vyper on August 28, 2002, 05:37:04 am
Btw,
Windows 2000
ICQ = 109386620
Post by: vyper on August 28, 2002, 06:11:29 am
Even in dos, I cannot manipulate the file. It is listed when you get a directory listing but DOS always kicks up "file not found" when u try to do anything to it.
Btw, sorry I lost you on icq sandwich - my dial up sucks today
Post by: Nico on August 28, 2002, 06:59:19 am
Quote
Originally posted by vyper
Well to update:
Even in dos, I cannot manipulate the file. It is listed when you get a directory listing but DOS always kicks up "file not found" when u try to do anything to it.
Btw, sorry I lost you on icq sandwich - my dial up sucks today
btw, by DOS, I mean reboot in dos mode, not just opening a dos windows. Is that what you did?
Post by: vyper on August 28, 2002, 07:56:09 am
Quote
Originally posted by venom2506
btw, by DOS, I mean reboot in dos mode, not just opening a dos windows. Is that what you did?
Yeh, I did that. I'm trying to tackle the problem through hex editing the hard drive, does anyone know of a way to stop win2k to stop you from changing things?
I'm using the latest version of Hackman
Post by: CP5670 on August 28, 2002, 08:42:30 am
Quote
Of course nobody else who posted on this thread never did, right?
What do you think in my case? :D I get warez all the time, but porn is for retards. :p
Quote
Bah, the entire situation has comedic value.
:lol:
Post by: TheCelestialOne on August 28, 2002, 08:56:02 am
Quote
Originally posted by CP5670
What do you think in my case? :D I get warez all the time, but porn is for retards. :p
So... CP... Why are you blushing?
Post by: CP5670 on August 28, 2002, 09:12:02 am
Post by: TheCelestialOne on August 28, 2002, 09:23:36 am
d u(x,t)/dt = C * d^2 u(x,t)/dx^2
on a 1D mesh from 0 to 1, with constant conductivity C, zero boundary conditions u(1,t)=u(0,t)=0, mesh spacing h=1/n, time step k, and z=k/h^2. We let U(i,m) denote our approximation of u(i*h,m*k).
Explicit solution of the 1D Heat Equation
for m = 1 to final m
for i = 1 to n-1
U(i,m+1) = z * U(i-1,m) + (1-2*z) * U(i,m) + z * U(i+1,m)
end for
end for
Let norm(U(:,m)) = max_i |U(i,m)|. We make two claims:
When 0 < z <= .5, norm(U(:,m+1)) <= norm(U(:,m))...
-----------
I won't continue but r u happy CP now that you can see some math? :D
Post by: Stealth on August 28, 2002, 09:27:53 am
restart in dos and manually delete it. if that doesn't work then format :D
Post by: CP5670 on August 28, 2002, 10:00:49 am
Anyway I think I found out what was wrong before; see, what I was trying to do was to derive euler's integral definition of the gamma function from weierstrass's product definition. When I expanded out the euler g constant in the zegz part into its limit definition, limk®¥ ( åkn=1 1/n - òk0 1/n dn ) and tried to merge it into the actual infinite product (this can be done by partitioning all the components of the exponent sum into eaeb... and so on), I forgot the last z for some stupid reason and got e1/z instead of k1/z, which screwed up the next part when I tried to convert the k into an appropriate finite product. No wonder... :rolleyes: :p
Post by: TheCelestialOne on August 28, 2002, 10:43:44 am
------------
We consider the following explicit algorithm for solving the heat equation
d u(x,t)/dt = C * d^2 u(x,t)/dx^2
on a 1D mesh from 0 to 1, with constant conductivity C, zero boundary conditions u(1,t)=u(0,t)=0, mesh spacing h=1/n, time step k, and z=k/h^2. We let U(i,m) denote our approximation of u(i*h,m*k).
Explicit solution of the 1D Heat Equation
for m = 1 to final m
for i = 1 to n-1
U(i,m+1) = z * U(i-1,m) + (1-2*z) * U(i,m) + z * U(i+1,m)
end for
end for
Let norm(U(:,m)) = max_i |U(i,m)|. We make two claims:
When 0 < z <= .5, norm(U(:,m+1)) <= norm(U(:,m)). In other words, the maximum absolute temperature can only decrease.
When z > .5, for any eps>0 there exists some n and some initial data U(:,0) with norm(U(:,0)) < eps such that norm(U(:,m)) becomes arbitrarily large as m grows.
In case 1, the algorithm is called stable; stability is necessary to get an accurate, physically meaningful result. In case 2, the algorithm is called unstable; instability yields an inaccurate and physically meaningless result.
Proof of Case 1. When 0 < z <= .5, then 0 <= 1-2*z < 1, so
|U(i,m+1)| <= z*|U(i-1,m)| + (1-2*z)*|U(i,m)| + z*|U(i+1,m)|
<= z*norm(U(:,m)) + (1-2*z)*norm(U(:,m)) + z*norm(U(:,m))
= norm(U(:,m))
Since this is true for all i, the result follows. This completes the proof of Case 1.
Proof of Case 2. Let U(i,0)=eps*sin((n-1)*i*pi/n), so norm(U(:,0)) <= eps. Then a little bit of trigonometry shows that
U(i,1) = eps*sin((n-1)*i*pi/n)* ( 1+2*z*(cos((n-1)*pi/n)-1) )
= U(i,0) * ( 1+2*z*(cos((n-1)*pi/n)-1) )
After m iterations, we get
U(i,m) = U(i,0) * ( 1-2*z + 2*z*cos((n-1)*pi/n) )^m
so
norm(U(:,m)) = norm(U(:,0)) * ( 1-2*z + 2*z*cos((n-1)*pi/n) )^m
The proof will be complete if we show that
1-2*z + 2*z*cos((n-1)*pi/n) < -1
since at each stage the solution U(:,m) will be multiplied by a number exceeding 1 in absolute value, and so grow without bound. After a little algebra, we get
1 - cos((n-1)*pi/n) > 1/z
Since z > .5, 1/z < 2. As n grows, cos((n-1)*pi/n) approaches cos(pi) = -1, so 1 - cos((n-1)*pi/n) approaches 2, and so eventually exceeds 1/z. This completes the proof of Case 2.
The background behind the Proof of Case 2 is this. We can express the inner loop of the algorithm in one line as
U(:,m+1) = T(z) * U(:,m)
where U(:,m+1) and U(:,m) are (n-1)-by-1 vectors, and T(z) is an (n-1)-by-(n-1) tridiagonal matrix:
[ 1-2*z z ]
[ z 1-2*z z ]
T(z) = [ ... ]
[ z 1-2*z z ]
[ z 1-2*z ]
One can use trigonometry to confirm that the eigenvalues and eigenvectors of T(z), T(z)*x(i) = lambda(i)*x(i), are given by
lambda(i) = (1-2*z) + 2*z*cos(i*pi/n)
and
x(i,j) = sqrt(2/n) * sin(i*j*pi/n)
The sqrt(2/n) factor guarantees that each eigenvector has unit length. In case 2, the largest eigenvalue in absolute value is lambda(n-1). We chose U(:,0) proportional to the corresponding eigenvector x(n-1), so at each step of the algorithm, U(:,m) just gets multiplied by the eigenvalue lambda(n-1).
Post by: Blue Lion on August 28, 2002, 10:49:10 am
This was better when it was about pr0n
Post by: CP5670 on August 28, 2002, 10:54:48 am
Are you into the methods of differential equations? Check out differential equation theory; very cool stuff. ;7
Post by: TheCelestialOne on August 28, 2002, 11:01:50 am
Quote
Originally posted by CP5670
but that's not the general solution! :p Also, I was talking the nD version, given by Ѳu(x1,x2,…) = 1/k ¶u/¶t. Try that one. :D (solution can be expressed as a fractional integral over n variables or a lauricella hypergeometric function)
What I just did is enough for one day... :sigh: As I said... I'm tired...
Post by: Blue Lion on August 28, 2002, 11:06:24 am
Post by: vyper on August 28, 2002, 11:34:15 am
any1 know of a free disk editor (hex, whatever - so long as I can find the reference for this file and nuke it)
Post by: Stealth on August 28, 2002, 11:37:01 am
Quote
Originally posted by CP5670
but that's not the general solution! :p Also, I was talking the nD version, given by Ѳu(x1,x2,…) = 1/k ¶u/¶t. Try that one. :D (solution can be expressed as a fractional integral over n variables or a lauricella hypergeometric function)
Are you into the methods of differential equations? Check out differential equation theory; very cool stuff. ;7
how can a 14 year old be this smart?
Post by: CP5670 on August 28, 2002, 12:01:06 pm
Quote
What I just did is enough for one day... As I said... I'm tired...
ah come on...how can you get tired of math? :p Alright, let's try another one then; how about a proof of the euler reflection formula, G(z)G(1-z)=pcsc(px), valid for all zÎC when z¹{1,0,-1,-2,...}. We will start with the expression of the left side:
G(z)G(1-z)
By the gamma functional identity, this can be written as
-zG(z)G(-z)
Expanding this into the weierstrass product definitions, we have:
-z×(zegz Õ¥k=1 (1+z/k)×e-z/k)-1×(-ze-gz Õ¥k=1 (1-z/k)×ez/k)-1
The factors egz, e-gz cancel out each other, as do the factors e-z/k and ez/k. The two z's can be multiplied to get z², and the negative signs cancel out. Since the order in which the terms in the products are multiplied does not matter, we can merge the two products into one:
z×(z² Õ¥k=1 (1+z/k)×(1-z/k) )-1
One of the pairs of z's outside the product cancels out, and the quantity inside the product can be expanded to get:
(-z Õ¥k=1 (1-z²/k²) )-1
Now let u=pz, so that z=u/p. Substituting this in, we have:
(u/p Õ¥k=1 (1-u²/(k²p²) )-1
Notice that the product combined with the u on the outside is now exactly that of the sine function sin(u), and this can be verified by taking the logarithm of the product to transform into a sum and then exponentiating. Therefore this becomes:
(1/p sin(u) )-1
We now substitute z back in and simply the expression to get the final result:
pcsc(pz)
This completes the proof. Beautiful, isn't it?
I would put up something else too but I have to go at the moment, so see you in a little while.
Post by: Knight Templar on August 28, 2002, 02:00:31 pm
Post by: Styxx on August 28, 2002, 02:05:06 pm
(and the offending post deleted or severely editted, of course)
Post by: Knight Templar on August 28, 2002, 02:15:22 pm
HARD CORE NUDITY
Post by: delta_7890 on August 28, 2002, 02:38:01 pm
Quote
Originally posted by Stealth
how can a 14 year old be this smart?
Perhaps being dropped as an infant had a reverse effect on him. Oo;;
Post by: Dr.Zer0 on August 28, 2002, 04:40:26 pm
Quote
Originally posted by Stealth
how can a 14 year old be this smart?
he's not human rember, he's a robot
Quote
Originally posted by delta_7890
Perhaps being dropped as an infant had a reverse effect on him. Oo;;
maybe the right side of his brain was damaged when he was droped leaving him with a fully functional left brain
Post by: TheCelestialOne on August 28, 2002, 04:42:32 pm
Quote
Originally posted by Dr.Zer0
he's not human rember, he's a robot
Aaaahhh! That explains his desire for a computerized government! ;)
Post by: penguin on August 28, 2002, 05:00:31 pm
Quote
Originally posted by Blue Lion:lol: :lol:
For the love of god...
This was better when it was about pr0n
Post by: Sandwich on August 28, 2002, 05:01:20 pm
Quote
Originally posted by Styxx
One more post about math here and this gets locked. ;)
(and the offending post deleted or severely editted, of course)
I'll race ya! :D
Post by: CP5670 on August 28, 2002, 07:00:12 pm
Quote
how can a 14 year old be this smart?
Actually my 16th birthday will occur in a week, but I thought everyone knows at least some of this stuff...
Quote
good lord cp, i'm not even going to try to understand that.
:wtf: eh...it wasn't that hard...
Post by: Stryke 9 on August 28, 2002, 07:13:16 pm
Actually, less porn viewing in connection with the art forum might be desirable. People might be able to see my renders if they didn't have geek love juice on their screens
Post by: Knight Templar on August 28, 2002, 11:13:51 pm