How do epsilon delta proofs work in math?


I have been trying to understand this part of math for a while, and no matter what it just seems very confusing.

In: Mathematics

2 Answers

Anonymous 0 Comments

Say you own a factory that purchases x’s and manufactures them into y’s.

The people buying your y’s call one day and say “Listen, there was too much error in this last batch of y’s. We need you to stay within [epsilon] of your target from now on, okay?”

So you call up your supplier and say “Hey, we are trying to run a tighter ship here and the incoming x’s seem to be an issue. Can you keep them within [delta] of the target from now on? Then the y’s coming out should be close enough for our new expectations.”

An epsilon-delta proof is figuring out how much error tolerance you need to demand from your suppliers (delta), based on how much is expected of you from your buyers (epsilon). If you can come up with a rule to pick an appropriate delta for any epsilon the suppliers may throw at you, then you are in good shape. Your factory can produce arbitrarily accurate y’s, given sufficient accuracy in the incoming x’s. We would say your factory is continuous (at that particular x value).

Anonymous 0 Comments

If a number is less than the minimum of two real numbers then it is less than the two simultaneously.

Let ϵ>0ϵ>0

If δ=min(1,ϵ/4)δ=min(1,ϵ/4) then for all xx such |x−1|<δ|x−1|<δ we have

|x−1|<1|x−1|<1 and |x−1|<ϵ/4|x−1|<ϵ/4

However |x−1|<1|x−1|<1 implies that |x|−1<1|x|−1<1 (because |x|−|1|≤|x−1||x|−|1|≤|x−1|) so |x|<2|x|<2 and then |x+2|<4|x+2|<4 (because |x+2|≤|x|+|2||x+2|≤|x|+|2|)

Finally |(x2+x+1)−3|=|x−1||x+2|<ϵ/4×4|(x2+x+1)−3|=|x−1||x+2|<ϵ/4×4