/sci/ - Science & Math

[Anonymous No.10274943]

epsilon delta breaks my brain

[Anonymous No.10274991]

>>10274943
it's not that hard to understand

norman wildberger probably has the most intuitive explanation for it that i've seen (even though he disagrees with it)

https://youtu.be/K4eAyn-oK4M?t=326

[Anonymous No.10274997]

>>10274943
>if the x's are close, then the f(x)'s are close
how exactly is this hard to understand
>>10274991
wildberger has impressively good lectures for someone so fucking nuts

[Anonymous No.10275009]

>>10274997
or maybe he's the only sane man in a nutty world.

[Anonymous No.10275012]

>>10275009
I'm sure were all absolute crazies for allowing dedekind cuts. Sure.

[Anonymous No.10275032]

>>10274997
i understand the concept and reading the examples makes sense but doing the exercises is really hard

[Anonymous No.10275123]

>>10275032
What are some exercises you're struggling with?
If your function has bounded slope (also called a Lipschitz function) then it's really simple. Delta is just epsilon over that slope. Think of it like a rectangle of height 2*epsilon you want to make thin enough to fit stuff through the sides.
If your function doesnt have bounded slope (like x^2) then split into an annoying region where it does, and a nice region where it doesnt. So we just look at the stuff outside of -1 to 1, since inside of this range we have bounded slope.
And out here we can deal with it in another way. Draw a picture of the new situation and see what delta works.
This is how you do any e-d, split into easy and tough situations and deal with each by drawing pictures to find the right delta.

[Anonymous No.10275192]

>>10275123
Right now we are only doing limits of sequences.
One of them is to prove the limit of (2n^2+3)/(n^2-n-cos(n)) is 2 as n tends towards infinity.
We haven't even proven shit like you can add/multiply limits yet.

[Anonymous No.10275241]

>>10275192
That's obnoxious. Here's how I'd do it.
Note that the top is greater than 2n^2 and the bottom less than n^2, so the whole thing is always greater than 2.
I'd look "when is (2n^2 + 3)/(n^2 - n - cos(n)) - 2 < epsilon?
Well, when something bigger than it is less than epsilon. Since the cosine is bounded below by -1, we can replace it by -1 and get something bigger (remember, smaller denom means bigger result)
So now we want (2n^2 + 3)/(n^2 - n - 1) - 2 < epsilon. Now you can solve for n if you want and you get what N should be.
I'll go through it but it shouldn't be too hard.
I'm noticing that it would be nice to get rid of the 2n^2, and I can. Let's rewrite 2:
(2n^2 + 3)/(n^2 - n - 1) - (2n^2 - 2n - 2)/(n^2 - n - 1) = (2n + 5)/(n^2 - n - 1) < epsilon.
Now for big n, n^2 - n - 1 > (n^2)/2, say n > 10. And 2n + 5 < 3n, say n > 5. So for sizeable n, (2n + 5)/(n^2 - n - 1) < 3n/(n^2/2) = 6/n, so now we only need that 6/n < epsilon.
Great! Now as long as we pick N > 6/epsilon, and N > 10 (we used that earlier) then for any n bigger than N,
(2n^2 + 3)/(n^2 - n - cos(n)) - 2 < 6/n < epsilon.

It's all about picking things to bound by other things, simplifying, and then picking more things to bound. Make it simpler and simpler by picking larger and larger bounds. Then as long as your biggest bound is less than epsilon, so will your original expression. And picking N for that big, simple bound should be easier!

[Anonymous No.10275266]

>>10274997
The nuclear method is to copy the proof for quotient, product sum whatever of limits with your particular function. That is horrific and tedious, and except your teacher is a cunt, you should always start and play with inequalities. It looks tough but you should not something when starting with [math] | \frac{2n^2+3}{n^2-n-cos(n)}-2|=|\frac{2n+2cos(n)+3}{n^2-n-cos(n)}| [/math] you get a nice lineal numerator now apply triangle inequality to separate it and you bound it by [math]\frac{2|n+cos(n)|}{|n^2-n-cos(n)|}+\frac{3}{|n^2-n-cos(n)|} [/math] And while you shouldn't take it as a law, when you have trig functions, plenty of times is a good idea to bound them by 1, because the analytical properties of such functions are difficult to work with in a rigorous setting so you can bound it even more, by another triangle inequality with [math]\frac{2(n+1)}{|n^2-n-cos(n)|}+\frac{3}{|n^2-n-cos(n)|} [/math] Now the tricky part is the denominator, but it isn't that tricky. The idea here is that because all quadratic functions increase at the same rate, you can bound it by another, more suitable quadratic. But now you are looking at things that are less than the denominator to bound it when taking reciprocals to completely forget about the cosine, you can take [math]n^2-n-2[/math] and because you can take an arbitrarily large N, you can consider only the values for which it is positive to forget about the absolute value and you get [math]\frac{2(n+1)}{n^2-n-2}+\frac{3}{n^2-n-2}[/math] now the second fraction can be bounded by 3/n because a quadartic always beats a linear, and for the second one find a constant "c" sufficiently large such that n^2-n-2>cn^2 and so you can get stuff that goes like 1/n. The complete procedure writing all out with proper notation is the disgusting stuff, but welp everyone has to do it. So because I want to larp as a professor, I leave that to you.

[Anonymous No.10275277]

>>10275266
As a side note, whenever there is a step when they tell you you can bound something by another thing for sufficiently large something, you should be making a list of what that N ism like [math]N_{i}[/math] then for the conclusion, you just take an N that is larger than the maximum of your set.

[Anonymous No.10275301]

>>10275266
>>10275241
Note how similar my (the first) and the second explanations are. Refocus your problem solving strategy from "I need to find N or delta so that this is less than epsilon" to "i need to find something easier that is bigger than this" and you'll have a much easier time with these proofs.

[Anonymous No.10275314]

>>10275301
Thank you anon
Ur a real human bean