Anyone in math?

f(t) + g(t) =< sup f(t) + sup g(t)

sup ( f(t) + g(t) ) =< sup ( sup f(t) + sup g(t) ) = sup f(t) + sup g(t)


no ?

i'm rusty :<

Say S = sup f(t)+g(t), F = sup f(t), G = sup g(t). Note that F > f(t), G > g(t) for all t in A so that F+G is an upper bound for f+g, but if S > F+G, the property of the supremum implies F+G is not an upper bound for f+g.

the key idea is that supremum is the least upper bound, so once you find an upper bound, you know the supremum has to be <= to it. thats basically what boblion did too.

heh..that was easy -_-...ya im rusty too. any examples of functions where equality doesnt hold? equality should hold for all continuous functions i think

im pretty sure it's true for all real functions. the proof assumes no special properties about f or g.

edit: misread your question, so ignore this

On April 09 2009 16:24 killThemDonks wrote: heh..that was easy -_-...ya im rusty too. any examples of functions where equality doesnt hold?

No idea but i guess it is always true.

On April 09 2009 16:24 killThemDonks wrote: equality should hold for all continuous functions i think

For sure.

are any of u in pure math? i may have some tougher questions..not homework or anything...just a refresher

On April 09 2009 16:35 killThemDonks wrote: are any of u in pure math? i may have some tougher questions..not homework or anything...just a refresher

I'm math/eco but i'm bad and rusty

yeah I am, just finished a class in real analysis last semester actually

hmm wait....equality holds for continuous functions in general....or continuous functions over a closed set...?? bah...soo rusty

i'm a math major.

equality holds

no wait...im an idiot...brb

now im confused...cuz say f, and g are continuous over some (closed) set T,
=> sup{ f(t): t in T} = max{ f(t): t in T} ...same holds for g(t)

Let h(t) = f(t) + g(t), it follows that h(t) is also continuous over T ==> sup h(t) = max h(t) = max( f(t) + g(t) ) =/= max( f(t) ) + max ( g(t) )

:S.. amirite?


take functions N(0,1) and N(1,1) for example over T = [-2,2]

so equality doesnt hold for cts functions????


math is idiotic.....

lol math

On April 09 2009 17:14 killThemDonks wrote: now im confused...cuz say f, and g are continuous over some (closed) set T, => sup{ f(t): t in T} = max{ f(t): t in T} ...same holds for g(t) Let h(t) = f(t) + g(t), it follows that h(t) is also continuous over T ==> sup h(t) = max h(t) = max( f(t) + g(t) ) =/= max( f(t) ) + max ( g(t) ) :S.. amirite? take functions N(0,1) and N(1,1) for example over T = [-2,2] so equality doesnt hold for cts functions???? math is idiotic.....

halp!

yeah, equality definitely doesn't hold even for continuous functions in general - it's true only if the maxes of f and g coincide at the same point

here's an easy one

f(t) = cos t
g(t) = -cos t

individually, each one oscillates between 1 and -1, thus each has a supremum of 1

add them together, and you get f(t) + g(t) = 0. the supremum is 0

zing

that is equality doesn't hold for those bad boys

that is sup ( f(t) + g(t) ) = sup ( 0 ) = 0 < sup f(t) + sup g(t) = 1+1 = 2

0 < 2

eh?

yups!

On April 09 2009 16:31 k3rn3l wrote:

I lol'd

On April 09 2009 17:58 Day[9] wrote: that is sup ( f(t) + g(t) ) = sup ( 0 ) = 0 < sup f(t) + sup g(t) = 1+1 = 2 0 < 2 eh?

:
So what ?
o,o
the equality holds o,o

Or i'm really tired, stupid and Romanian.

Day u Fail

^ equality doesnt hold necessarily for continuous functions (not sure why i thought so in the first place :S).

i need sleep maybe ?

omg

i'm retarded. I thought Day said that the whole "=<" was wrong lol

day9 is smart
and right

how did i receive a degree in mathematics O___o

E=Mc2

On April 09 2009 18:12 NiTE wrote: Show nested quote + On April 09 2009 16:31 k3rn3l wrote: I lol'd

LMFAO this is the first thing I thought when I saw the equation. Well, not exactly a cat saying "sup" but definitely the "Sup?" mentality came to mind.

Wish I could help, I haven't touched any math for 3 years and this looks beyond anything I knew even when I was doing this back in high school. I miss math...