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Inequality with modulus

I don't even know how to start. :frown:
(edited 9 years ago)
Reply 1
This is essentially an extension of the usual triangle inequality. This is one of the few times I advocate thinking about things geometrically (:lol:)

You can think of the LHS as a 'straight line' from x to y

The RHS is the journey: x to a, a to b, b to y.

The latter is only ever going to be as short as the straight line x to y if a and b lie on the straight line from x to y.

(Note, the algebraic proof to this is also straight forward, but thinking about it geometrically can help motivate how to start)
(edited 9 years ago)
Reply 2
If that isn't helpful, the key hint is to use the triangle inequality.
Original post by Noble.
If that isn't helpful, the key hint is to use the triangle inequality.


Is it possible with vectors?


Posted from TSR Mobile
Reply 4
Original post by rayquaza17
I don't even know how to start. :frown:


|x-y|=|x-a+a-b+b-y|=|(x-a) + (a-b) + (b-y)| <= .... by a famous result
Reply 5
Original post by TeeEm
|x-y|=|x-a+a-b+b-y|=|(x-a) + (a-b) + (b-y)| <= .... by a famous result


What's the famous result?

Thanks, I had written down (x+a)+(b-a)-(y+b) [after seeing Noble's hint] and I was wondering why it didn't work.

Edit: I've now realised I had it written down like that because I had written the question down wrong.:facepalm:

Original post by Noble.
If that isn't helpful, the key hint is to use the triangle inequality.


Nope, it was helpful! Thank you!

I know it was such a simple question, but it's one of the first ones on a long list of analysis questions I need to do and I think because it's on that list that my mind is going blank. (Analysis and me don't really get on.)
(edited 9 years ago)
Reply 6
Original post by physicsmaths
Is it possible with vectors?


Posted from TSR Mobile


Yes, you can do it with Rn\mathbb{R}^n as well, from the same geometric principal.
Reply 7
Original post by rayquaza17
What's the famous result?

Thanks, I had written down (x+a)+(b-a)-(y+b) [after seeing Noble's hint] and I was wondering why it didn't work.



Nope, it was helpful! Thank you!

I know it was such a simple question, but it's one of the first ones on a long list of analysis questions I need to do and I think because it's on that list that my mind is going blank. (Analysis and me don't really get on.)


You can do it his way, or you can approach it from the other end

xa+ab+byxb+byxy|x-a| + |a-b| + |b-y| \geq |x-b| + |b-y| \geq |x-y|

With the first step applying the triangle inequality to the first two terms, then the second step applying it again.
Reply 8
Ok this is a different question, but I think it's the same sort of method:


Would this be right:
xnx_n converges with limit l means, given any ϵ2>0\frac{\epsilon }{2}>0, there is NNN\in \mathbb{N} such that xnl<ϵ2|x_n-l|<\frac{\epsilon }{2} for all nNn\geq N

Then we can write lxm<ϵ2|l-x_m|<\frac{\epsilon }{2}

Then:
xnxm=xnl+lxmxnl+xxm<ϵ2+ϵ2=ϵ|x_n-x_m|=|x_n-l+l-x_m|\leq |x_n-l|+|x-x_m|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon
Reply 9
Yes. What you've shown is that a convergent sequence is Cauchy. Just to nitpick, the question should say xnxm<ϵ|x_n - x_m| < \epsilon for all n,mNn,m \geq N, not just n.

You can try the harder direction as well, that every Cauchy sequence is convergent.

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