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#math
<Cpudan80> Let me mathbin the question
<gzl> no. then you have to figure out for what values of x_1, x_2, x_3, x_4 you get that all entries in Ax are zero.
<Cpudan80> hrm
<JIMJONESBALLIN> wb trwbw.
<Cpudan80> gzl: You mind if I mathbin the question, and my idea to solve the problem?
<gzl> I'm not going to look at it, but you're welcome to for someone else
<TRWBW> Cpudan80: you better check on #math-permission-to-ask first
<TRWBW> Cpudan80: but gzl has a point, if you type it in ascii more people will consider it
<cerealkiller219> umm...
<cerealkiller219> Question: I'm solving for the derivative and I'm down to (x + 1) d/dx (1/x^(2/3))
<Olathe> The derivative of what ?
<cerealkiller219> well, 4(x+1) / (3x^(2/3))
<cerealkiller219> but im down to that section ^^
<Olathe> Alright.
<Olathe> 1/x^a = x^-a
<TRWBW> cerealkiller219: hint, 1/u^(2/3)=u^(-2/3)
<cerealkiller219> i know that but... the answer they show is : 2(x+1) / 3(x^(5/3))
<cerealkiller219> wwhich isnt at all what I've calculated.
<cerealkiller219> -2(x+1) **
<Cpudan80> Here's the question:
<Cpudan80> http://www.mathbin.net/8231
<TRWBW> cerealkiller219: d/dx x^a=a*x^(a-1). tht hold for a negative too.
<TRWBW> cerealkiller219: i don't know why they don't have a - sign, but otherwise
<cerealkiller219> I know the rules of derivatives, I just dont know how they simplified their answer in that manner
<cerealkiller219> yeah I'm stumped =(
<TRWBW> Cpudan80: that's more complicated than i want to explain. it's not deep, but the sum of a bunch of simple details
<Cpudan80> TRWBW: Can you just augment the matrix so the last column is zeros and solve it?
<TRWBW> Cpudan80: that works
<hefty> augment, that reminds me of a really nasty antibiotic i use to take when i was a little kid
<Cpudan80> that would be augmentin
<Cpudan80> and yes - it's a nasty powder form of penicillin
<hefty> yes it would
<hefty> in liquid its HORRIBLE
<TRWBW> Cpudan80: or you could solve for the first 3 equalling the last column, then your general solution is the a,b,c you get for the first 3 columns, with -1 for the last column, times anything
<TRWBW> Cpudan80: assuming the first 3 are linearly independnet, i can't tell by eyeballing
<TRWBW> btw, if you ever go into a pet store they sell the basic antibiotics, penicicllin, tetracycline, over the counter for animals. dirt cheep. just so you know.
<hefty> interesting
<hefty> what cant you get over the counter or on the black market these days
<hefty> you can get everything from like car parts to just shy of body parts
<TRWBW> hefty: well on the black market i guess you can get armor piercing bullets full of cocain and heroin and dipped in plutonium.
<hefty> excellent
<Cpudan80> I wouldn't use the animal antibiotics
<Cpudan80> They are not FDA approved for human consumption
<hefty> yeah that probably isnt safe lol
<TRWBW> Cpudan80: i haven't, but i would. that's just silly. it would be a lot more expensive for them to make different for the animals. i'd bet 10 to 1 they come from the same factory that your doctor prescribed ones come from.
<hefty> good point
<Kampen> man, why is the #c channel invite only >_<
<Kampen> i wanted someone to look over my code to make sure there isn't anything else i can do to reduce the time complexity. but i guess i can't.
<TRWBW> Kampen: i can't remember why i'm banned form #freenode-social. i could only speculate, sometimes i'm blunt and unpolitically correct, i must have been there last year.
<Golfgeo> Hi all, got a question: I'm using Mathcad to solve some MTTF equesions and would like to transform the working worksheet into one function. But there doesn't seem to be an option for this???
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