#math - Sun 15 Apr 2007 between 00:03 and 00:42

NY Lost Funds



lesshastewell ,,, I changed the lines to 3 320 120 55.92 (i.e. I removed the names)
and I did scan(file = "scores.txt")
but it doesn't seem to have loaded it in in a useful way
anyone able to help
?
TRWBWlesshaste: this isn't really the channel for that. it's more for the theory than how to use a particular program. try something like #stats i'm guessing.
lesshastesadly no such channel
TRWBWabout as likely to get your answer here as in #klingon-speakers
lesshasteoh
R is quite popular amongst mathematicians
WILDSTYLER is alright.
lesshastecan I ask a question about the coupon collector's problem?
WILDSTYLEI'm kinda partial to C.
lesshasteWILDSTYLE: is that a joke?
TRWBWi tried R, but when back to using C and Fortran
JabberWalkielesshaste: no
you have to read the topic first
TRWBWlesshaste: no, for statistics it's a lot nicer to do it in a real programming language. it runs faster, you can get libraries with better algorithms, etc.
Zanco-afIRCHello Everybody
WILDSTYLEhello.
jadenbaneDoes Sum (1, inf., 1/(n(n+1))) = 0 ?
Mulderjadenbane,
no
for starters, 1/2 is one of the terms you add
so automatically, your sum is > 0
jadenbaneHaha, I should actually look at the numbers shouldn't I :P
Sum (1, inf., 1/(n(n+1))) = Sum (1, inf., 1/n - 1/(n+1)) which makes me think there will be cancelation.
Mulderif you can decompose it into a partial fraction lik ethat
you can telescope, and work out what the infinite sum is
and and your final sum is probably something like 1
jadenbaneSee, I'm not sure what telescoping is, since I only have a vague notion of ir.
*it.
Mulderwell, a series is just a seqeuence of partial fractions
jadenbaneOh, that sum will give 1/1 - 1/2 + 1/2 + 1/3 - 1/3 etc.
Mulderyes
jadenbaneSo yeah, I was hasty in guessing zero. I should have guessed one.
Mulderso s_n = 1 - 1/2 + 1/2 + 1/3 - 1/3 + ... - 1/(n-1) + 1/(n-1) - 1/n
s_n = 1 + 1/n
then if we take the limit of that, we get 1
since 1/n goes to 0 as n -> Infinity

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NY Lost Funds