General Math Discussion Topic (not a forum gaaame)

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General Math Discussion Topic (not a forum gaaame)

Post by Blast!10 » Wed Jul 14, 2010 4:41 pm

This topic is to discuss anything and everything math related. Talk about conversion formulas, fun facts, proof, bitwise operators, algebra, and everything else.

Did you know that there's one easy way to find a conversion formula between Celsius and other temperature scales?

First of all, if someone presents you a new scale, he probably tells you the freezing and boiling point of water. In Celsius it's 0 and 100, and in Fahrenheit 32 and 212.

I'll make an example by converting Celsius to Fahrenheit. Celsius will be the "source scale" and Fahrenheit will be the "destination scale".
The first thing you need to do is subtract the destination scale (F) freezing temp (of water) from the destination scale boiling temp. (212-32=180)
Divide the result by [source boiling point]-[source freezing point]. If you'd do Fahrenheit to Celsius, you would divide the result by 212-32=180. Since I'm converting from Celsius, I'm dividing the result by 100-0=100. 180/100=1.8, which means that each degree Celsius is 1.8 degrees Fahrenheit. Of course if you already know the distances between each degree, you can skip this step entirely.

The next step is very simple. If you're converting from Celsius, multiply the source temperature by the "distance" between each degree in the destination scale (in my case 1.8) and add the freezing point of water in the destination scale (in my case 32).

If you're converting to Celsius, subtract the freezing point of water in the destination scale from the source temperature, and divide the result by the "distance" between each degree in the destination scale.

Let's say my temperature was 40 Celsius. In Fahrenheit that would be 40*1.8+32=104.

To convert 104 Fahrenheit back to Celsius, do (104-32)/1.8=40.

Dlcs18 once made a scale called Unfriad, where water boils at 9001 and freezes at -9001. I've found that each degree Celsius is 180.02 degrees Unfriad, and made Celsius-Unfriad and Unfriad-Celsius conversion formulae, all using the method described above.

27 Celsius in Unfriad would be 25*180.02+(-9001)=25*180.02-9001=4140.46.

This isn't very understandable, but I did my best. My country isn't an English-speaking one, so I'm quite unfamiliar with English math terms. :roll: :wink:

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Post by DEEMAN223344 » Wed Jul 14, 2010 4:47 pm

IT'S OVER NINETHOUSAAAAAAND!!!
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Post by ~xpr'd~ » Wed Jul 14, 2010 4:54 pm

g(x)=x^4+2x^3+x^2+8x-12

g(x) = x^4 + 2x^3 + x^2 + 8x - 12

g(x) = (x + 3)(x - 1)(x^2 + 4)

Real zeros are x = -3 and x = 1

The other 2 complex roots are ±2i.

Yet I'm not totally sure :E


and i also suck at writing math
nobody understands mine
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Post by dlcs18 » Wed Jul 14, 2010 4:55 pm

EDIT: This is actually useless because it just produces x*y*z. Nothing special.

If you have 3 integers, named x, y, and z, and do this formula with it:
((x*y)/z)*((x*z)/y)*((y*z)/x), you will always get a whole number.

I've tried it with: (number in brackets is the result)
2 3 4 (24)
4 5 6 (120)
6 7 8 (336)
11 12 13 (1716)

I think the reason it's always a whole number is because you are using the same three numbers every time you do (blah*blah)/blah, so the fractions merge together to make a whole number.
Last edited by dlcs18 on Wed Jul 14, 2010 5:23 pm, edited 3 times in total.
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Post by Blast!10 » Wed Jul 14, 2010 5:02 pm

Just checked, it works with 125 126 127.

Also, I think you missed a bracket there.
((x*y)/z)*(x*z)/y)*((y*z)/x)
The big bracket has no opening bracket. Is this a mistake?

EDIT: I just checked, and found another error. ((4*5)/6)*((4*6)/5)*((5*6)/4)=120, not 12.
Last edited by Blast!10 on Wed Jul 14, 2010 5:06 pm, edited 1 time in total.
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Post by dlcs18 » Wed Jul 14, 2010 5:05 pm

yes... fix'd
EDIT: Oh, ok. I guess that explains why at first the results seemed to be going up and down. Fix'd that too. :lol:
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Post by MyNameIsKooky » Wed Jul 14, 2010 5:14 pm

dlcs18 wrote:If you have 3 integers, named x, y, and z, and do this formula with it:
((x*y)/z)*((x*z)/y)*((y*z)/x), you will always get a whole number.
If you multiply the denominators with denominators and numerators with neumerators you'll get:
(x^2*y^2*z^2)/(xyz). Dividing those leaves you with xyz and thus it's not a fraction at all, so you'll always get whole numbers.

EDIT: Unless you plug in fractions. :P
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Post by dlcs18 » Wed Jul 14, 2010 5:19 pm

Oh... How could I have not realised it was just producing x*y*z? I fail.
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Post by Blast!10 » Thu Jul 15, 2010 2:29 pm

^_^ wrote:I hate word "White".
Its annoying when you say that like 1000000000000 times.
Let's see, how much time will it take my computer to write the word "white" 1 trillion times?

This is the PHP code I used:

Code: Select all

<?

$i=0;
$myFile = "lol.txt";
$fh = fopen&#40;$myFile, 'a'&#41;;

$Data = "white ";
while &#40;$i < 1000000000000&#41; &#123;
fwrite&#40;$fh, $Data&#41;;
$i++;
&#125;

fclose&#40;$fh&#41;;

?>
Using that code, it took my computer one minute to write "white" 167,209 times (see this link). Dividing that by 60 makes 2,786.8166... which is roughly 2,787. This means my computer writes 2787 copies of the word "white" per second.
If I want to find out how many seconds my computer will spend writing "white" 1 trillion times I have to divide 1 trillion by 2,787. That makes roughly 358,808,755. 358,808,755 seconds = 5,980,146 minutes = 99,669 hours = 4,153 days = about 138 months = 11 years.

Nope, I'm not going to wait 11 years for my computer to do that.

Also, thanks to tyteen for helping me with the code.
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Post by Marinus » Thu Jul 15, 2010 3:51 pm

Blast!10 wrote:
^_^ wrote:Its annoying when you say that like 1000000000000 times.
Nope, I'm not going to wait 11 years for my computer to do that.
And then your computer has not even said the word, only written on the screen. Let's say a computer can say the word at most 4 times/second (while you still can understand it) it will take about 8000 years. Yes, I guess that will be pretty annoying. :lol:
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Post by LittleZbot » Thu Jul 15, 2010 9:38 pm

*sees this is about Math*

*sighs*

*runs off*

I hate Math.

*runs farther*

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To Marinus, who was my community older brother.
To Janet, who I will remember every time I wear a toque.

May these Wonderlanders find more adventures in passing than they ever had in life.
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Post by Marinus » Thu Jul 15, 2010 9:44 pm

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Last edited by Marinus on Fri Jul 16, 2010 12:07 am, edited 1 time in total.
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Post by MyNameIsKooky » Thu Jul 15, 2010 10:01 pm

A few months ago, I thought up a nice formula for finding square numbers one at a time. Think about these square numbers...

1
4
9
16
25

Each number is 2n+1 after each other where n is the number's square root. Here's the formula applied in the space after.

1 +3 [1+2(1)+1=3, we add the answer to the original number to get the next square number, 4]
4 +5
9 +7
16 +9
25 +11
...

Thus, we can find that the next square number after 25 is 36, which has a square root of 6, and so on. Continuing the sequence using the formula to help...

36 +13
49 +15
64 +17
81 +19
100 +21
121 +23
144 +25
169 +27
Etc...

:D
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Post by Marinus » Thu Jul 15, 2010 10:16 pm

I think you're right Kooky, but I see an even easier pattern: Each next square is the previous square + the next odd number (not quite sure if odd is the right word but I mean a number not to be divided by 2 and still an integer): 1, 3, 5, 7, etc...

Edit: But I have to admit that yours is more Mathematical: (n+1)^2 = n^2 + 2n + 1
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Post by Nobody » Thu Jul 15, 2010 11:35 pm

BEEF!

What is it? :twisted:
i should change my signature to not be rude to pcpuzzle
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Post by MyNameIsKooky » Thu Jul 15, 2010 11:42 pm

BEEF = 48879

:twisted:
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Post by Marinus » Thu Jul 15, 2010 11:48 pm

I did not calculate it, but I guess it's hexathings. :lol:
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Post by mqdar » Fri Jul 16, 2010 10:44 am

This isn't directly related to maths, but it makes no sense to me and I can't find any info about it with on the internet.

Have any of you ever opened WA level files in hex-editors? Some numbers are stored in a really strange way. Here's four different ones I've found:

c0 3f = 1
20 40 = 2
60 40 = 3
90 40 = 4

Double-digit hex values for storing simple numbers. Help please? :?
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Post by Marinus » Fri Jul 16, 2010 11:28 am

Probably it doesn't make sense at all, but I found a pattern. If you write those numbers binary, you get this:

1100 0000 0011 1111
0010 0000 0100 0000
0110 0000 0100 0000
1001 0000 0100 0000

When you take only the 2nd to 9th bits (blue ones) and mirror them, you get 1, 10, 11 and 100, which is 1,2,3 and 4. The other bits (black ones) don't make any sense to me; maybe they are used for something else.

But maybe it's complete nonsense though. :roll:
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Post by MyNameIsKooky » Fri Jul 16, 2010 4:51 pm

The hex numbers are either pointlessly compressed or encrypted. But what Marinus said seems a lot more likely, and seems to point towards the fact that the hex numbers in the level files are encrypted.
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