#math - Thu 22 Feb 2007 between 00:57 and 01:59

arrenlexTRWBW: Yeah, A=(1/(b-a))integral(f,a,b). But I can't figure out how to use that.
me22Abs[min value of function] <= (integral over width) / width <= Abs[max value of function]
TRWBWarrenlex: if f(x)<=c then integral is <= (b-a)*c. if f(x0)<c then |f(x)-f(x0)|<(c-f(x0))/2 for x in some |x-x0|<delta.
Kampenif anyone is able, could they look up and answer my question about the notation for tychonoff spaces?
TRWBWarrenlex: integrate [a,x0-delta] [x0-delta,x0+delta] [x0+delta,b] and add them up
arrenlex: you'll get the average is less than the average, proof by contradiction
arrenlex: anotherway, use g(x)=f(x)-(average of f(x)). same problem, but it might be clearer to you see how you can show that if g(x)<=0 and for some x0 g(x)<0, then integral g(x)<0
arrenlexI've considered g(x)=A-f(x). I worked out that integral(g)=0 because integral(A)=A(b-a)=((b-a)/(b-a))int(f)=int(f) and int(f)-int(f)=0.
Does that relate to what you're suggesting?
(A is the average value of f)
Or is my logic wrong?
TRWBWarrenlex: it does, but that doesn't make it much easier, they key point is that if a continuos function drops below the average it must drop below it for some delta around where it drops. then you can break up the integral, like i said. read my posts.
arrenlexI'm trying to understand them, yeah. Thanks a lot for your help.
TRWBWarrenlex: np, gl
arrenlex: maybe this will help you see it, maybe not. if you have a finite sum a_1+..+a_n, every a_i is less than equal to the average, can any of the a_i be less the the average?
arrenlexTRWBW: Thanks a lot for trying, but I think I'm getting somewhere with what's been said already; thanks lots, though.
TRWBWarrenlex: with the trick of subtracting the average, that becomes: if you have sum a_1+...+a_n, a_i<=0, and at least one of the a_i<0, can they sum to 0?
seb-_TRWBW: no
TRWBWseb-_: <arrenlex> *screams* this seems SO OBVIOUS but I'm completely stuck! How do you prove that a continuous function that never exceeds its average value must be constant?
seb-_TRWBW: heh
arrenlex: i hope you got it now
arrenlexYep, I got it. Thank you, all.
seb-_arrenlex: what was the mental snag?
arrenlexMe not understanding that A-f(x)>0.
>=0, sorry.
That means that g(x)=A-f(x)>=0, but since integral(g)=0 as I said above, that means g must be 0 for the whole interval, because it's non-negative.
Which means that A=f(x) for all x.
cerealkiller219seb: you nearby
thermoplyaeIn case you guys had any doubts, set theory is hard
arctanxo \in{\mathbb{R}}?
Kasasdkadthermoplyae: I had no doubts
Though I doubt it's as hard as classical plane geometry
Which is pretty much the most difficult thing ever
thermoplyaePretty much
Kampenwhat, specifically, is the field that deals with topological groups called?
a professor said that isn't algebraic topology
buti don't know what else it could be
thermoplyaeI think plane geometry and number theory are the two big towers of mathematics that I will never really venture into
I often see them in the distance, but...
Kampenis it more just algebra?

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