TriggerX
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April 23, 2015, 06:26:55 AM Last edit: April 23, 2015, 06:37:58 AM by TriggerX |
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I think I have the answer, can you escrow the funds? Also do you think you can extend the date a few more days?
Edit: Nevermind, I'll just post the answers. Only for part a)
alpha = (mu)/[(mu+L)(mu+3L)] xi = 3L/u c = (s_k)(u+L)(u+L)(u+3L)(u+3L)/3L k = (1/[(s_k)/m-(1+1/xi)*c*alpha*alpha*2*L)] + (mu+L)/[2*alpha*mu*L]
If it is correct, please send the BTC to this address: 13MLwx49jT4a2SSwD17qu1uaG1sbGMdrzi
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Hi!
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oxiyusuf
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April 23, 2015, 07:30:13 AM |
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Im still confuse with math question, for this one, very difficult for me! Im usually can answer for aljabar question!
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erikalui
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April 23, 2015, 10:31:52 AM |
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Checking the solution!
Am I correct that you are suggesting the following?
k=1
alpha=(u + L)/2uL
e =4uL - 1
c=(sk/m - 1)/(1 + 1/(4uL - 1))*2u²L/(u + L)²
EDIT: I still need to check it, but I don't think it will work, since your first three lines contain a mistake... 1/k - 2alpha*u*L/(u + L) <=0; 1/k <= 2alpha*uL/(u + L); k<=(u + L)/2alpha*uL
EDIT2: This solution does not work, since for u = 1, L =2, the second (original) equation does not hold. (1+4*1*2-1)*(3)/(2*1*2) - 2/(1+2) = 5.3333... which is not <=0
I just checked and noticed I substituted a wrong value for alpha and so the result is incorrect. I can see that another memebr has tried solving the equation and if by chance his solution too is incorrect, I can try again.
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Agestorzrxx
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April 23, 2015, 12:30:17 PM |
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you can use Mathematica to solve this problem, it's not hard.
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arrowguys
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April 23, 2015, 03:03:44 PM |
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the answer is n ^ n + b ^ = c n ^
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Baldassare (OP)
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April 23, 2015, 07:40:18 PM |
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TriggerX,
Let u=1/100, L=2/100, s=1, m=2
Then xi = 6, c = 147/2000000, a = 100/21.
In that case the first three inequalities hold, but the last one doesn't.
BOUNTY STILL OPEN!!!!!
EDIT: I will pay $20 for EITHER a) or b) !!!! Will extend the deadline 24 hours from now.
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TriggerX
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April 24, 2015, 01:23:46 AM Last edit: April 24, 2015, 02:42:57 AM by TriggerX |
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xi = 2L/mu alpha = 1/(1+xi)(mu+L) c = (s_k)/(2*xi*mu*m*alpha*alpha) k=(1/[(s_k)/m-(1+1/xi)*c*alpha*alpha*2*L)] + (mu+L)/[2*alpha*mu*L]
Is this correct?
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Hi!
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Baldassare (OP)
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April 24, 2015, 04:24:39 AM |
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TriggerX, this looks extremely close! Still need to carefully check
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ShetKid
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April 24, 2015, 05:58:26 AM |
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I am not Math genius, but in high school we were told that if we had 5 variables and 4 equations for it, then it wouldn't be unsolvable . Wouldn't that be true for this case ?
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monsterdoge
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April 24, 2015, 06:22:31 AM |
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math make me crazy insane lol , goodluck to everyone .
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bitspill
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April 24, 2015, 07:57:27 AM |
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wouldn't be unsolvable
May want to rephrase that a bit
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volatilebtc
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April 24, 2015, 03:40:10 PM |
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where is the fund being escrowed ? Show us the proof ?
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Baldassare (OP)
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April 24, 2015, 04:01:53 PM |
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TriggerX did not request an escrow. I'm checking his solution at this moment. If it's good, I'll send the money.
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monsterdoge
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April 24, 2015, 04:21:07 PM |
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so triggerx already won this contest?
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Baldassare (OP)
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April 24, 2015, 04:55:31 PM |
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Yes, TriggerX got an answer. Thanks everyone! Contest closed.
TriggerX, money sent. Would you mind posting the steps to the solution?
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TriggerX
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April 25, 2015, 01:17:48 AM |
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Yes, TriggerX got an answer. Thanks everyone! Contest closed.
TriggerX, money sent. Would you mind posting the steps to the solution?
It took 30 minutes to figure it out and I don't want to type it all out again haha. If I have time I may post it here again. As for the BTC, I have received it, thanks.
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Hi!
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