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#math
<yi> ok, will do, thanks.
<yi> ok, something more concrete now. Find f(x), a power series in x, such that (1+x)f`(x) = mf(x), where m is not in the natural numbers.
<A-MART> me22, n^2007 = 4014
<yi> how do you cook something like that up?
<Olathe> O-o
<Kasasdkad> yi: why not just solve the differential equation for general m?
<yi> because I don't remember differential equations at all :)
<me22> A-MART: check that, I think you forgot something
<A-MART> er...
<me22> 3^4015 * 3^-1 is most certainly *not* equal to 4014 :P
<A-MART> 3^4015 * 3^-1 = 3^(4015 - 1) = 3^4014
<JIMJONESBALLIN> heh.
<A-MART> my bad
<me22> A-MART: so if it were m^4014 < 3^4014, could you do it?
<TRWBW> yi: if that's a power series, that's gonna give you a relation on the coefficients a_i
<TRWBW> yi: btw, if you just want one solution, try f(x)=0. with power series 0+0+0+... then (1+x)*0=m*0 for any m. 1, pi, sqrt(2), whatever
<A-MART> :o
<A-MART> 8
<Olathe> 9
<A-MART> 9 would be equal to it, right?
<A-MART> er..
<Kasasdkad> TRWBW: I think he died
<topbloke> @arcsin[1/2]
<mbot> Unknown command, try @list
<Olathe> % ArcSin[1/2]
<mbot> Olathe: Pi/6
<topbloke> o nice
<hedos> % d[x^2/2]/dt
<TRWBW> Kasasdkad: i like giving people the answer to their problem after they quit irc ;)
<mbot> hedos: d[x^2/2]/dt
<Dacicus> You're trying to differentiate x^2/2 wrt t?
<hedos> % d[t^2/2]/dt
<mbot> hedos: d[t^2/2]/dt
<Olathe> % D[t^2/2, t]
<mbot> Olathe: t
<hedos> Oh :)
<hedos> @list
<mbot> http://www.cse.unsw.edu.au/~dons/lambdabot/COMMANDS
<TRWBW> is there a way i can make my chat client filter out all lines that start with %?
<hedos> %list
<Riastradh> That depends on what client you're using, TRWBW.
<Olathe> % help
<mbot> Olathe: "See http://documents.wolfram.com/v5/ for detailed Mathematica help."
<hedos> thanks
<Olathe> You're welcome.
<Riastradh> I've no idea how to do it with mIRC, though, short of reading the documentation.
<A-MART> The number 8 is the sum and product of the numbers in the collection of four positive integers {1,1,2,4} since 1+1+2+4 = 8 and 1*1*2*4 = 8. The number 2007 can be made up from a collection of n positive integers that multiply to 2007 and add to 2007. What is the smallest value of n with n>1?
<TRWBW> A-MART: start by factoring 2007
<Dacicus> yeah
<A-MART> okay
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