#math - Sun 1 Apr 2007 between 01:35 and 01:42

NY Lost Funds



TRWBWmingkus: um, nope
mindwarp: it's f(x)=log(1+x), f'(x)=1/(1+x)
futurist0 and ln2
TRWBWfuturist: f'(x) not f(x)
futuristoh, 1/2 and 1
mingkusyou are trying to prove this right x/2 < log(1+x) < x ??
TRWBWfuturist: and if using the mean value you get x*f'(t) for some t in [0,x] which means t also in [0,1], what can you say about x*f'(t)?
futurist: if f'(t) is in [1/2,1], what can you say x*f'(t) is in?
futuristlemme think about that for a second
TRWBWfuturist: if a is in [1/2,1], what can you say a*b is in?
futuristoh [b/2,b] -- i was starting at the x*f'(t) thing
mingkusfuturist: x/2 < log(1+x) < x for x in [0,1], is that what you want to prove?
TRWBWfuturist: okay, so back to the last one
futurist: if f'(t) is in [1/2,1], what can you say x*f'(t) is in?
futuristmingkus yeah
oh, [x/2,x]
hmm weird
TRWBWfuturist: look helpful?
mingkusfuturist: is it log(1 + x) or ln(1 + x)
SafroleHow about using the convexity of the log function?
Is that applicable to this problem?
futuristso x/(1+x) is in the desired intervale
TRWBW?
futuristf'(x) = 1/(1+x)
mingkusnot if it log
futuristi mean, x/(1+x) is in the interval we want to show log(1+x) is in
ln
TRWBWfuturist: you want to show that f(x) is in [x/2,x]. MVT tells you f(x)=f'(t)*x for some t in [0,x]. you just showed f'(t)*x is in [x/2,x]
futuristlog = ln in this book
mingkuslog at base e is equal to ln
k i know how to do it
futuristmingkus yes
actionTRWBW shaves his head
TRWBWshaves his head
oops shakes
mingkusdevide in 2...
Safrolelol
mingkusx/2 < log(1+x) and log(1+x) < x
TRWBWfuturist: you want to show that f(x) is in [x/2,x]. MVT tells you f(x)=f'(t)*x for some t in [0,x]. you just showed f'(t)*x is in [x/2,x]
mingkusor ln sorry
futuristah

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NY Lost Funds