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# Choosing epsilon for a sequence that converges watch

1. Problem + Solution
Definition for convergence

I'm really confused on how epsilon is chosen (i.e. the red arrows in the picture), and how those inequalities are worked out.

In this example, is there a simpler value of epsilon that could be chosen?

I (think I) understand how it works - you're proving that once you reach a certain point in the sequence (N) |an-l|, the difference between a term in the sequence and its limit, will always be smaller than an arbitrarily chosen small number.

But I'm still lost on how epsilon is chosen.

Many thanks.
2. ε can be any ( small number )... for instance if someone says ε = 0.05 then you put that in to the formula...

so if N is greater than or equal to √(9/2*0.05) then |an - 0.5| < 0.05 as required
3. (Original post by the bear)
ε can be any ( small number )... for instance if someone says ε = 0.05 then you put that in to the formula...

so if N is greater than or equal to √(9/2*0.05) then |an - 0.5| < 0.05 as required
I think I understand all of that, but when you're solving it for yourself, how do you know to choose √(9/2ε)
I think I understand all of that, but when you're solving it for yourself, how do you know to choose √(9/2ε)
you just need to look at half a dozen worked problems to get the theme of finding epsilon
5. (Original post by the bear)
you just need to look at half a dozen worked problems to get the theme of finding epsilon
ok, in this example is there anything else fairly obvious that epsilon could have been, or is this the only real solution?
ok, in this example is there anything else fairly obvious that epsilon could have been, or is this the only real solution?
i am good with that version

https://bobobobo.wordpress.com/2008/...year-calculus/
ok, in this example is there anything else fairly obvious that epsilon could have been, or is this the only real solution?
You seem a little confused; your aim is find N given a particular value of .

So I suspect you mean is there anything else obvous that N could be (other than .

And the answer is yes. All you need to do is find an N that works (i.e.N such that for every n > N we have ). So obviously if N works, anything bigger than N will work too.

so would work, for example.

Note in particular that there's no requirement for N to be the smallest value that will work. It's important to understand this, since It's often impossibly hard to find the smallest value that will work.
9. (Original post by DFranklin)
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(Original post by the bear)
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Thanks for the replies!
edit: so many mistakes in this it was embarassing.
edit wow I messed that one up - here's my uncertain-corrected solution, which I'm hoping some can check

http://s21.postimg.org/8gxp59rfr/20151026_203856.jpg
Your solution (both of them!) seem to make the claim that , which is clearly nonsense.

This is a "fatal" mistake, there is nothing recoverable from what you've done after that point.

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Updated: October 26, 2015
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