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#1 2006-02-22 17:56:28
- rimi
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- Registered: 2006-02-21
- Posts: 17
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.
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#2 2006-02-22 20:07:09
- MathsIsFun
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- Registered: 2005-01-21
- Posts: 7,713
Re: Try this one
To do superscripts:
u[sup]n[/sup]and you get:
u[sup]n[/sup]
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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#3 2006-02-22 20:32:17
- Jai Ganesh
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- Registered: 2005-06-28
- Posts: 48,422
Re: Try this one
Hence, the question would be.....
Prove that u[sup]n[/sup] x v[sup]1-n[/sup]≤nu+(1-n)v
where
n∈(o,1) and u,v > 0
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#4 2006-02-23 04:51:50
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
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 u[sup]n[/sup] x v[sup]1-n[/sup] ≤ 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.
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#5 2006-02-24 17:28:07
- rimi
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- Registered: 2006-02-21
- Posts: 17
Re: Try this one
No mathsy thats not quite a proof...but I can give you hints if you want...this is simple...indeed
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#6 2006-02-28 17:35:07
- rimi
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- Registered: 2006-02-21
- Posts: 17
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.
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#7 2006-02-28 19:00:37
- krassi_holmz
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- Registered: 2005-12-02
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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.
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#8 2006-03-01 17:35:24
- krassi_holmz
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- Registered: 2005-12-02
- Posts: 1,905
Re: Try this one
Rimiq please, help us more.
IPBLE: Increasing Performance By Lowering Expectations.
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