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81  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: February 07, 2015, 11:02:01 PM
I'm not entirely sure what you're asking here OP, but a couple of things popped into my mind while RE-reading:

1. Using the formula for new_x i presented, if we have to start with x = 9, then we have (x2 - 1)2 = 6400, which means the formula will never reach 10 for any p where 6400 = 0 (mod p), or for which (64002 - 1)2 = 1677721518080001 = 0 (mod p), etc.
 
2. Though you possibly qualified it by saying "at least in the context of...", I just thought I'd note that 2 is not necessarily a generator of ℤp× where p is prime. Consider, for example, p = 7.

Of course, I may be misunderstanding entirely what you're wanting to do.
82  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: February 07, 2015, 08:21:42 PM
can you please give me exactly what you are looking for in a CLEAR AND PRECISE PRivate Message? cause i mean you were unclear to begin with and every post its all jumping around to different maths and shit. like seriously you want me to solve thois shit you gotta be clear what you are looking for what types of maths are involved and what type of algorithm or mathematical formula you are requesting. i only went off of what your OP stated (which, was still very unclear)
83  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: February 07, 2015, 08:14:47 PM
WHAT IS WRONG WITH WHAT I POSTED?! WHAT IS NOT CORRECT?! YOU POSTED SOMETHING YOU NEEDED AN ANSWER FOR, I DID THE MATH AND THE ALGORITHM, IT IS NOT INCORRECT!!!!
84  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: February 07, 2015, 07:20:14 PM
I HAVE SOLVED THE ISSUE WITH THE MATHEMATICAL EQUATION! ITS FINALLY CLEAR NOW!!!!
here
var equation = equation.replace("x", "(x²-1)² / (4*x*(x²+a*x+1))");
should read
var equation = equation.replace(/x/gi, "(x²-1)² / (4*x*(x²+a*x+1))");
which I already tested for you
<script>
var equation= "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))";
var a=0;var b=1;var x=9;

document.write(equation+"</br>");
while(a<b)
{
var equation = equation.replace(/x/gi, "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))");

document.write(equation+"</br>");
a=a+1;
}
var x2=x*x;


var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));

document.write(answer+"</br>");
var answer=((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");

var x=9;

var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");
var x=answer;
var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");



var x=9;
var answer=((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1)*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1) / (4*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+a*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+1))
document.write(answer+"</br>");
</script>
I also changed (x²-1)² / (4*x*(x²+a*x+1)) and used (x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1)) instead because javascript can directly evaluate it. Just for demonstation purposes I left
var x=9;

var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");
var x=answer;
var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");
which is evaluation by recursion. Find the answer plug it back into the equation etc. Compared with the created equation
var x=9;
var answer=((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1)*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1) / (4*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+a*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+1))
document.write(answer+"</br>");
both of which return the same answer so the code in the end is
<script>
var equation= "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))";var a=0;var b=6;var x=9;var c=0;document.write(eval(equation)+"</br>");
while(c<b){var equation = equation.replace(/x/gi, "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))");
document.write(equation+"</br></br>");document.write(eval(equation)+"</br>");c=c+1;}
</script>
This will give you just the answers minus the equations
<script>
var equation= "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))";var a=0;var b=6;var x=9;var c=0;document.write(eval(equation)+"</br>");
while(c<b){var equation = equation.replace(/x/gi, "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))");
document.write(eval(equation)+"</br>");c=c+1;}
</script>




Please send bounty to THIS BTC adress!!!
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
85  Economy / Services / Re: Need PHP Coder to edit my store on: February 07, 2015, 07:13:29 PM
Since the project was not completed. NET5kY keep silent and no reply to my skype pm. And I paid 500$ to get a not complete source?
How do people think?
So fun

ive decided to take this work, simple enough, user paid deposit ($100.00) even though i said it was not necessary. i appreciate it though and i will continue work now.

Bid: $1500.00USD
Deposit: $100.00USD
Remaining Balance: 1400.00USD
86  Bitcoin / Bitcoin Discussion / Crypto laws? on: February 04, 2015, 10:52:35 PM
seeing as Bitcoin is starting to gain traction in the legal sense all over the world, why dont we just ditch bitcoin and keep moving onto other cryptos? cause in reality they wont wanna keep having to push new paperwork because of the name legality (bitcoin,paycoin,etc). i think they will just finally give up on it entirely and then we can make a real crypto and prevent the whole mining difficulty issue we have now with bitcoin. i think it makes perfect sense to keep switching so local govts get tired of resubmitting paperwork for different named coins, cause they have to be specific in paperwork.
87  Economy / Services / Re: Looking for experienced programmer - must know PHP, mySQL, JavaScript on: January 30, 2015, 04:15:13 PM
im interested please hit me up on skype or pme me here my skype is pythonpro1337 thanks
88  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 06:27:07 PM
THE FORMULA YOU REQUIRE IS CALLED  pemdas
89  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 06:42:20 AM
""The winner is the first person to post such formula in private. The formula must work in all cases, and be computationally feasible (let us say calculatable in less than 24 hours)""

Well simply put, i did this, but with php and html lol the formula is written in a script in a private message in his inbox. i hope he gets on son cause the anticipation is killing me
90  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 06:28:27 AM
its funny cause the script ive written and sent him like 5 hours ago wil display the correct answer using the algorithm that i used i like how everyone is trying to step on my fucking toes cause i KNOW the one i sent him is correct. OP please make the right decision here as to who gets the bounty cause ive sent multiple pm's thanks again
91  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 04:15:47 AM
anyone know where the original poster is at?!
92  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 03:46:59 AM
i tried and came up with this.. hope i can get some xD

=((x-1)^2 (x+1)^2)\/(4 x (a x+x^2+1))


=(x^4-2 x^2+1)\/(4 a x^2+4 x^3+4 x)


x = 1, a - 2=/0
x = 1, a + 2=/0

={x element R : (a<=-2 and x<0) or (a<=-2 and x>0 and sqrt(a^2-4)+a+2 x<0)
or (a<=-2 and sqrt(a^2-4)+a+2 x>0 and sqrt(a^2-4)>a+2 x)
or (a<=-2 and a+2 x>sqrt(a^2-4))
or (-2<a<2 and x!=0)
or (a>=2 and sqrt(a^2-4)>a+2 x and sqrt(a^2-4)+a+2 x>0)
or (a>=2 and sqrt(a^2-4)<a+2 x and x<0)
or (a>=2 and x>0)
or (a>=2 and sqrt(a^2-4)+a+2 x<0)}

1\/(4 x)-a\/4+1\/4 (a^2-3) x+(a-a^3\/4) x^2+1\/4 (a^4-5 a^2+4) x^3+O(x^4)\n(Laurent series)


x\/4-a\/4+(a^2-3)\/(4 x)+(a-a^3\/4)\/x^2+O((1\/x)^3)\n

(d)\/(dx)((x^2-1)^2\/(4 x (x^2+a x+1))) = ((x^2-1) (2 a (x^3+x)+x^4+6 x^2+1))\/(4 x^2 (a x+x^2+1)^2)

 integral (-1+x^2)^2\/(4 x (1+a x+x^2)) dx = 1\/8 ((a^2-4) log(a x+x^2+1)+2 a sqrt(4-a^2) tan^(-1)((a+2 x)\/sqrt(4-a^2))-2 a x+x^2+2 log(x))+constant



BTC: 199hWWjMZdZ59dfyKpi7AG7wKv9LnqJSij



i tried and came up with this.. hope i can get some xD

=((x-1)^2 (x+1)^2)\/(4 x (a x+x^2+1))


=(x^4-2 x^2+1)\/(4 a x^2+4 x^3+4 x)


x = 1, a - 2=/0
x = 1, a + 2=/0

={x element R : (a<=-2 and x<0) or (a<=-2 and x>0 and sqrt(a^2-4)+a+2 x<0)
or (a<=-2 and sqrt(a^2-4)+a+2 x>0 and sqrt(a^2-4)>a+2 x)
or (a<=-2 and a+2 x>sqrt(a^2-4))
or (-2<a<2 and x!=0)
or (a>=2 and sqrt(a^2-4)>a+2 x and sqrt(a^2-4)+a+2 x>0)
or (a>=2 and sqrt(a^2-4)<a+2 x and x<0)
or (a>=2 and x>0)
or (a>=2 and sqrt(a^2-4)+a+2 x<0)}

1\/(4 x)-a\/4+1\/4 (a^2-3) x+(a-a^3\/4) x^2+1\/4 (a^4-5 a^2+4) x^3+O(x^4)\n(Laurent series)


x\/4-a\/4+(a^2-3)\/(4 x)+(a-a^3\/4)\/x^2+O((1\/x)^3)\n

(d)\/(dx)((x^2-1)^2\/(4 x (x^2+a x+1))) = ((x^2-1) (2 a (x^3+x)+x^4+6 x^2+1))\/(4 x^2 (a x+x^2+1)^2)

 integral (-1+x^2)^2\/(4 x (1+a x+x^2)) dx = 1\/8 ((a^2-4) log(a x+x^2+1)+2 a sqrt(4-a^2) tan^(-1)((a+2 x)\/sqrt(4-a^2))-2 a x+x^2+2 log(x))+constant



BTC: 199hWWjMZdZ59dfyKpi7AG7wKv9LnqJSij



can check the times of his post my posts on the forums and messages in your inbox i messaged you first money please and thank you
93  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 03:39:55 AM
yeah ive given him the instructions to find his answer and how to do it in which languages thank you!
94  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 03:29:54 AM
so the bounty is @ 34 for me now? Wink please check your messages form me ive finished this bounty thanks!
95  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 02:55:50 AM
yes i just sent him how to solve it in pm
96  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 02:48:29 AM
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN

i just sent you a pm with the appropriate script on how to compute and find your solution please send my bounty the addy above thanks you!

@OP:You can run it yourself in a .html file just copy and paste it into notepad. Of course over 100000 is kinda slow on my computer so there are some limitations to it. If I was to have it calculated I would run the above script as php and then use that to generate the javascript code to give me an answer.

there your problem is solved in your pm please send me the bounty and close this thread
97  Economy / Services / Re: [30 BTC Bounty] - Find a mathematical / algorithmical formula on: January 29, 2015, 02:40:59 AM
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN

i just sent you a pm with the appropriate script on how to compute and find your solution please send my bounty the addy above thanks you!
98  Economy / Services / Re: [For Hire] on: January 27, 2015, 08:00:36 PM
ok 30% deposit upfront and remaining on milestones VIA escrow. thanks kindly everyone lookiong forward to working on projects with you!
99  Economy / Services / [For Hire] on: January 27, 2015, 04:02:31 PM
Hi everyone, im a developer looking for work. i can do websites and ecommerce as well as bitcoin payment integration and paid fantasty league websites for cash prizes and etc. i can also code in many linux dependencies including python ruby lua and more. message me for skype. i charge 35% deposit and milestone payments! thanks guys!
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