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#1 2006-02-23 16:56:28
- rimi
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Try this one
prove that ,
u (to the power)n × v(to the power)(1-n) ≤nu + (1-n)v
n∈(o,1) and u,v > 0
by the way how do I bring this superscripts ? As word files are not working.
#2 2006-02-23 19:07:09
- MathsIsFun
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Re: Try this one
To do superscripts:
Code:
u[sup]n[/sup]
and you get:
un
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#3 2006-02-23 19:32:17
- ganesh
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Re: Try this one
Hence, the question would be.....
Prove that un x v1-n≤nu+(1-n)v
where
n∈(o,1) and u,v > 0
Character is who you are when no one is looking.
#4 2006-02-24 03:51:50
- mathsyperson
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Re: Try this one
If n = 0, then the inequality simplifies down to v ≤ v.
Similarly, if n = 1, then the inequality simplifies down to u ≤ u.
So un x v1-n ≤ nu+(1-n)v all the time because they are actually always equal to each other.
Why did the vector cross the road?
It wanted to be normal.
#5 2006-02-25 16:28:07
- rimi
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Re: Try this one
No mathsy thats not quite a proof...but I can give you hints if you want...this is simple...indeed
#6 2006-03-01 16:35:07
- rimi
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Re: Try this one
I guess I can give you some hints about the solution...you can consider the fact that log is a concave function.
#7 2006-03-01 18:00:37
- krassi_holmz
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Re: Try this one
We must prove that
log(nu+(1-n)v)>=nlog(u)+(1-n)log(v)
Is there some unequation for
log(a+b) and log(a)+log(b)?
IPBLE: Increasing Performance By Lowering Expectations.
#8 2006-03-02 16:35:24
- krassi_holmz
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Re: Try this one
Rimiq please, help us more.
IPBLE: Increasing Performance By Lowering Expectations.