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#1 2006-02-22 17:56:28

rimi
Member
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
Administrator
Registered: 2005-01-21
Posts: 7,550

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

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,780

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


Character is who you are when no one is looking.

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#4 2006-02-23 04:51:50

mathsyperson
Moderator
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
Member
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
Member
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
Real Member
Registered: 2005-12-02
Posts: 1,908

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
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Try this one

Rimiq please, help us more.


IPBLE:  Increasing Performance By Lowering Expectations.

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