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March 19th, 2011, 06:58 AM   #1
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open ball

Let , with . If , prove that .
retep is offline  
March 19th, 2011, 07:23 AM   #2
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Re: open ball

We have 0<1-t<1 and by triangular inequality we have . Now you have to show that the inequality is strict.
girdav is offline  
March 20th, 2011, 06:03 PM   #3

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Re: open ball

Originally Posted by retep
Let , with . If , prove that .

Considering x and y complexes. If their moduli are the same, they are different only in phase(they can be equal too). Both x and y lie in a circle of radius r in the complex plane. (1-t)x+ty, , is the segment which connects y to x. |(1-t)x+ty|, , is the distance of a point in that segment(that is a circle's chord) to the complex plane origin, excluding x and y. You can, that way, see that |(1-t)x+ty|<r in an intuitive manner.

But intuition is not a proof, so:

By the triangular inequality:

, : inequality ( 1 )

to prove that the inequality is strict you can calculate the values of t which for the equality holds. Make and .

: expression ( i )

To hold, in expression ( i ) or . But and than, the inequality ( 1 ) is strict.

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