Hard Light Productions Forums

Off-Topic Discussion => General Discussion => Topic started by: HotSnoJ on March 24, 2003, 07:19:52 am

Title: check this out, 1MB host!
Post by: HotSnoJ on March 24, 2003, 07:19:52 am

http://www.clickherefree.com/cgi-bin/webdata_FreeWebHosting.pl?cgifunction=form&fid=1048029535


ROTFL:lol: :lol:

Title: check this out, 1MB host!
Post by: Ashrak on March 24, 2003, 07:29:35 am

oooh
:eek2: :wtf: :lol:

Title: check this out, 1MB host!
Post by: Ulundel on March 24, 2003, 07:36:14 am

Wowsers... :lol:

Title: check this out, 1MB host!
Post by: Tiara on March 24, 2003, 07:44:50 am

:eek2:

Think of the possibilities you will have with1 MB!:blah:

:lol:

Title: check this out, 1MB host!
Post by: Fetty on March 24, 2003, 07:59:40 am

Forced Ads: Banner/Top  ;7

Title: check this out, 1MB host!
Post by: Stealth on March 24, 2003, 10:35:12 am

i guess if you want a 2 page HTML website that's ok ;)

Title: check this out, 1MB host!
Post by: Carl on March 24, 2003, 10:51:02 am

though if you want to make an art gallery, this may be the place. you could fit a dozen jpgs or so in there.

Title: check this out, 1MB host!
Post by: Gortef on March 24, 2003, 11:28:52 am

w00t :D :lol:

Title: check this out, 1MB host!
Post by: Stealth on March 24, 2003, 11:48:08 am

Quote

Originally posted by Carl
though if you want to make an art gallery, this may be the place. you could fit a dozen jpgs or so in there.

yeah, a dozen 40x40 pixel JPEGs ;)

Title: check this out, 1MB host!
Post by: Petrarch of the VBB on March 24, 2003, 01:04:20 pm

What. Is. The. Point.

Title: check this out, 1MB host!
Post by: Tiara on March 24, 2003, 01:14:22 pm

Code: [Select]

dy/dx = ay - by^3
    dy/dx - ay = -by^3    Divide through by y^3
  (1/y^3)dy/dx) - a/y^2 = -b

Make the substitution u = 1/y^2  So du/dx = (-2/y^3)(dy/dx)
                                  (-1/2)du/dx = (1/y^3)dy/dx)

Substitute for y and dy/dx

     (-1/2)du/dx) - au = -b
           du/dx + 2au = 2b

Multiply by the integrating factor e^(INT(2a.dx))
                                = e^(2ax)

 e^(2ax).du/dx + 2au.e^(2ax) = 2b.e^(2ax)

    d/dx{u.e^(2ax)} = 2b.e^(2ax)

Integrate   u.e^(2ax) = 2b.INT{e^(2ax).dx}
            u.e^(2ax) = 2b(1/(2a)).e^(2ax) + const.

      (1/y^2).e^(2ax) = (b/a).e^(2ax) + const


Thats the point... :p Get it now? :wink:

Title: check this out, 1MB host!
Post by: Martinus on March 24, 2003, 01:23:43 pm

Quote

Originally posted by Tiara

Code: [Select]

dy/dx = ay - by^3
    dy/dx - ay = -by^3    Divide through by y^3
  (1/y^3)dy/dx) - a/y^2 = -b

Make the substitution u = 1/y^2  So du/dx = (-2/y^3)(dy/dx)
                                  (-1/2)du/dx = (1/y^3)dy/dx)

Substitute for y and dy/dx

     (-1/2)du/dx) - au = -b
           du/dx + 2au = 2b

Multiply by the integrating factor e^(INT(2a.dx))
                                = e^(2ax)

 e^(2ax).du/dx + 2au.e^(2ax) = 2b.e^(2ax)

    d/dx{u.e^(2ax)} = 2b.e^(2ax)

Integrate   u.e^(2ax) = 2b.INT{e^(2ax).dx}
            u.e^(2ax) = 2b(1/(2a)).e^(2ax) + const.

      (1/y^2).e^(2ax) = (b/a).e^(2ax) + const


Thats the point... :p Get it now? :wink: [/B]

[color=66ff00]Ok CP! What did you do with Tiara's body? :wtf:
[/color]

Title: check this out, 1MB host!
Post by: an0n on March 24, 2003, 02:15:12 pm

Methinks you phrased that wrong.

Title: check this out, 1MB host!
Post by: Martinus on March 24, 2003, 02:26:50 pm

Quote

Originally posted by an0n
Methinks you phrased that wrong.

[color=66ff00]Methinks you have a twisted imagination.

Considering CP's stance on girls how can you think of anything in that ballpark? :p
[/color]

Title: check this out, 1MB host!
Post by: WMCoolmon on March 24, 2003, 05:32:50 pm

Quote

This is getting out of hand...now there are two of them!

:nervous:

Title: check this out, 1MB host!
Post by: DragonClaw on March 24, 2003, 06:08:40 pm

Looks just like what I was looking for....  :blah:

Title: check this out, 1MB host!
Post by: Carl on March 24, 2003, 06:13:29 pm

Quote

Originally posted by Stealth


yeah, a dozen 40x40 pixel JPEGs ;)

you can get a 800x600 to 80kb on good compression.

Title: check this out, 1MB host!
Post by: Tiara on March 25, 2003, 02:41:29 am

...

Why is it that you guys think nonly CP knows his math?

Quote

z = tan(x/2).

Then

     sin(x) = 2*z/(1+z^2)
     cos(x) = (1-z^2)/(1+z^2)
         dx = 2*dz/(1+z^2)

Substitute this in, and you'll end up needing to integrate a rational function of z:

       INTEGRAL dx/(3+4*sin

• )^2

     = INTEGRAL 1/(3+4*2*z/[1+z^2])^2 * 2*dz/(1+z^2)
     = INTEGRAL (1+z^2)^2/(3*z^2+8*z+3)^2 * 2*dz/(1+z^2)
     = INTEGRAL 2*(1+z^2)/(3*z^2+8*z+3)^2 * dz

Followed by;

     3*z^2 + 8*z + 3 = 3*(z + [4+sqrt(7)]/3)*(z + [4-sqrt(7)]/3)

and partial fractions to break it into parts which you can easily
integrate.

Title: check this out, 1MB host!
Post by: CP5670 on March 25, 2003, 09:04:08 am

:wtf: what was the original problem in that last quote?

try this one: ;7

¥
ò ( e^(1-x)t - 1 ) log(t) / ( e^t - 1 ) dt
0

Title: check this out, 1MB host!
Post by: Tiara on March 25, 2003, 09:13:17 am

Quote

Originally posted by CP5670
:wtf: what was the original problem in that last quote?

Integrating Trig Function: dx/(3+4sin x)^2

Very simple I know... :p

Title: check this out, 1MB host!
Post by: Razor on March 25, 2003, 09:21:35 am

Quote

Originally posted by CP5670
:wtf: what was the original problem in that last quote?

try this one: ;7

¥
ò ( e^(1-x)t - 1 ) log(t) / ( e^t - 1 ) dt
0

You have 2 unknowns there. You must give us a value for X if you want us to solve that.

By the way, you are insane. How the heck do you integrate a function when it's deffinition is from 0 to unlimited?

EDIT: Here is how I started solving that:

(http://www.fattonys.com/images/upload/image1.jpg)

Title: check this out, 1MB host!
Post by: CP5670 on March 25, 2003, 09:40:20 am

actually the answer to that is a function of x. :D your image is a red x though.

Quote

By the way, you are insane. How the heck do you integrate a function when it's deffinition is from 0 to unlimited?

eh...take appropriate limits, how else?

Title: check this out, 1MB host!
Post by: Martinus on March 25, 2003, 12:31:45 pm

Quote

Originally posted by Tiara
...

Why is it that you guys think nonly CP knows his math?

 

[color=66ff00]:lol: It's usually only CP who throws a math problem up out of the blue, just for the hell of it.
We're more used to you running around 'axing' people and slitting the throats of those that notice your flagrant use of elven underwear. ;)
[/color]

Title: check this out, 1MB host!
Post by: Razor on March 25, 2003, 01:08:15 pm

Quote

Originally posted by CP5670
actually the answer to that is a function of x. :D

You are starting to scare me now. :nervous: :shaking:

Title: check this out, 1MB host!
Post by: CP5670 on March 25, 2003, 01:23:45 pm

alright I will give you the answer; see if you can prove why it is so. :D The integral equals log PF_-1(x) + g H(x) , where PF is the powerfactorial, H is the harmonic number and g is the euler constant; if x is a positive integer, PF_n(x+1) = 1¹ⁿ2²ⁿ3³ⁿ...x^xn and H(x+1) = 1/1+1/2 +1/3+...+1/x.

Title: check this out, 1MB host!
Post by: Galemp on March 25, 2003, 01:59:39 pm

I hate you.

Title: check this out, 1MB host!
Post by: SKYNET-011 on March 25, 2003, 02:50:33 pm

Arrrgh maths!

Title: check this out, 1MB host!
Post by: Razor on March 25, 2003, 04:39:38 pm

Quote

Originally posted by CP5670
alright I will give you the answer; see if you can prove why it is so. :D The integral equals log PF_-1(x) + g H(x) , where PF is the powerfactorial, H is the harmonic number and g is the euler constant; if x is a positive integer, PF_n(x+1) = 1¹ⁿ2²ⁿ3³ⁿ...x^xn and H(x+1) = 1/1+1/2 +1/3+...+1/x.

Yeah...eh...I get it. Jesus Christ! I am never gonna take another math course again.