l don't understand the way $\epsilon$ and $\delta$ are being used in this question about proving $|u-y|<\delta\Rightarrow|u^n-y^n|<\epsilon$

153 Views Asked by At

The full statement is: Given $y\in\mathbb{R},n\in\mathbb{N}$ and $\epsilon>0$, show that for some $\delta>0$, if $u\in\mathbb{R}$ and $|u-y|<\delta$ then $|u^n-y^n|<\epsilon$.

Ultimately, I'm not really sure how $\epsilon$ and $\delta$ are related to one another in this sense, and because of that I don't really get what this is asking me to prove. Am I just supposed to be showing theres some random real number $\delta$ that satisfies this? That seems like it would it be really trivial since theres no largest real number, so my thinking is that that would be the wrong interpretation. Any help on interpreting this would be greatly appreciated.

Lastly, I saw a question pertaining to this proof, but they skipped the base case, and don't really ever talk about what the proof means, so none of their reasoning made any sense. I don't really want help doing the proof, just some help with interpreting the question I guess.

1

There are 1 best solutions below

5
On BEST ANSWER

The full statement you must prove, with all of the quantifiers in the correct position, is this:

for all $y \in \mathbb R$, $n \in \mathbb N$, $\epsilon > 0$ there exists $\delta > 0$ such that for all $u \in \mathbb R$, if $|u-y| < \delta$ then $|u^n - y^n| < \epsilon$.

In this statement, the values of $y$, $n$ and $\epsilon$ are given to you (that's what always happens when you prove a "for all" statement).

Your job is to use the given values of $y$, $n$ and $\epsilon$ to find an appropriate value of $\delta > 0$ --- perhaps expressed as a formula in terms of $y$, $n$ and $\epsilon$ --- and then, using that value of $\delta$, to prove the following implication:

for all $u \in \mathbb R$, if $|u-y| < \delta$ then $|u^n - y^n| < \epsilon$.

You ask: "Am I just supposed to be showing there's some random real number $\delta$ that satisfies this?" Well, yes and no. You are supposed to be showing that there's some real number that satisfies this implication (subject to the requirement that the number be positive). But the real number you find will be far from random. That's what I meant by "appropriate". If you just pluck any old value of $\delta$ out of a hat, it may well be inappropriate: the implication may be false when using that value of $\delta$.

Another thing to keep in mind is that you are not asked to find a unique value of $\delta$. In fact, if you found one value of $\delta > 0$ which made that implication true, then any smaller value of $\delta$ would also make that implication true. But, you don't have to find all possible values of $\delta$ which make the implication true, you just have to find one.

You don't ask for a detailed proof, and I'm happy not to give one, but let me nonetheless point out what might be the key strategy: start with the inequality $|u^n - y^n| < \epsilon$ and attempt to "solve" it for the quantity $|u-y|$, obtaining in the end an inequality of the form $|u-y| < \text{SOMETHING}$ which, by a chain of backward implications, implies the desired inequality $|u^n - y^n| < \epsilon$; then you just set $\delta = \text{SOMETHING}$.

Finally, I have a suggestion: prove it first for $n=1$; and then for $n=2$. If you get that far, you will have learned the $\epsilon$-$\delta$ method. The rest is just icing.