the question and where i got stuck in this link:

http://img187.imageshack.us/img187/8868/44429492ix7.gif

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- Jan 6th 2009, 10:31 AMtransgalacticlimit prooving question..
the question and where i got stuck in this link:

http://img187.imageshack.us/img187/8868/44429492ix7.gif - Jan 6th 2009, 11:45 AMKrizalid
We have to prove the following statement:

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Observe that Now pick and

Hence, conditions are fulfilled if you take - Jan 6th 2009, 12:43 PMtransgalactic
my question i about how to choose the value for M

?? - Jan 6th 2009, 12:59 PMtransgalactic
i understand that our upper bound is 11

so inorder for this to work (like your previous examples in the article you published

we need only

why to use minimum??

the previoss werenot solved like that

- Jan 6th 2009, 01:18 PMtransgalactic
we get x as close as possible to 5

|x-5|<1 ==>> -1<x-5<1 so x+5<11

so M>=11

lambda=e/11

why minimum with 1 ?? - Jan 6th 2009, 03:17 PMKrizalid
It's just any number you want.

I took because is a nice guy. You could've taken other delta, a ratio, 2, whatever. - Jan 6th 2009, 09:20 PMtransgalactic
why you didnt do this min function in the previous examples

in the article? - Jan 6th 2009, 10:11 PMtransgalactic
usually ou ended this question by saying lambda=e/11

why lambda=min{1,e/11}

why 1? why any number i want?

why minimum? - Jan 7th 2009, 01:19 AMtransgalactic
did i prove it correctly

i solved it differently from the article

is that ok?

http://img224.imageshack.us/img224/9929/93418507so5.gif