EpsilonDelta proof
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EpsilonDelta proof
I'm going to be taking Calculus I this semester. Last semester I dropped the course early on because I couldn't get the hang of the EpsilonDelta proof. Could you guys give me your best understanding of the proof and how to do it?
Re: EpsilonDelta proof
DaPwnzlord wrote:I'm going to be taking Calculus I this semester. Last semester I dropped the course early on because I couldn't get the hang of the EpsilonDelta proof. Could you guys give me your best understanding of the proof and how to do it?
Think of it as a badguy good guy game. You are the good guy. The bad guy picks epsilon, you have to produce a delta (or show one exists). Since smaller epsilons are 'harder' (since they encompass all bigger epsilons), expect him to pick really small epsilons.
The most straightforward way to do this is to find an explicit delta in terms of epsilon, but you need to make sure that the delta you get is positive for all (small) epsilon.
So an example.
[math]\lim_{x\rightarrow 2} x^2 = 4[/math]
So, what do we need to show? If some badguy picks a very tiny but positive [imath]\varepsilon[/imath], we need to be able to find a positive [imath]\delta[/imath] so that
[math]x2< \delta \Rightarrow x^2  4 < \varepsilon[/math]
To work with this, we can assume [imath]x2<\delta[/imath], and we want to start with [imath]x^24[/imath] and try to bound it above.
So
[math]x^24=(x2)(x+2) = (x2)(x+2) < \deltax+2[/math]
Ok, this is good, because we now just need to make sure [imath]\deltax+2 < \varepsilon[/imath] and then we are done! But x is close to 2, so x+2 has to be pretty close to 4. Check that as long as we take delta less than 1, then x+2 is positive but less than 5. So now we have [math]x^24<5\delta.[/math] Pick [imath]\delta = min(\varepsilon/5,1)[/imath] and we are done!
Last edited by mikel on Wed Jan 20, 2010 3:08 pm UTC, edited 1 time in total.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
Re: EpsilonDelta proof
Also, read the definition again and again and again. Every time you read it, you make another copy, so just read it 'til it sticks. Think about it, try to visualize it. You'll lay in your bed one night and go "ah!". From there, it's a matter of practice. Talking to your class mates about it should help. Preferably someone who's already got it.
It's one of those things you'll get the hang of gradually over the semester, so don't get discouraged if you don't get it right away.
It's one of those things you'll get the hang of gradually over the semester, so don't get discouraged if you don't get it right away.
Re: EpsilonDelta proof
Also, don't drop Calc I even if you cannot understand epsilondelta proofs! The proofs are rather advanced, and in any Calc I class I've ever seen, much less than 5% of the material. You are not likely to see them again in the class, and unless you are a math major, won't see them again period. That's not to say that they're not worth understanding, but they're certainly not worth dropping calc I over.
Re: EpsilonDelta proof
Stickman wrote:You are not likely to see them again in the class, and unless you are a math major, won't see them again period.
Even then, it's not something that you'll use to establish limits or continuity of a particular, specific function like x^{2}. That's something which is rather silly. Why should I have to use epsilondelta to prove that a polynomial is continuous? :\
 Talith
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Re: EpsilonDelta proof
Proving that a polynomial in x is continuous for all x in R requires that you show that [math]\lim_{x \to c}P(x) = P(c), \forall c \in \mathbf{R}[/math] which you can only show by going from the original epsilon delta proofs. (Note that showing a finite sum of limits is equal to the limit of the finite sum requires an epsilondelta proof)
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Re: EpsilonDelta proof
How about the function definition of continuity/limits based of "modulus of continuity"?
In that variant, to show a function is continuous you have to produce a modulus of continuity (a function) that, when given the epsilon, produces the delta.
This is done explicitly, instead of explicitly, in the definition.
Ie:
[math]\lim_{x\rightarrow 2} f(x) = 4[/math]
means:
There exists a function u called "the modulus of continuity" such that for all [imath]\varepsilon[/imath] > 0, 0 <  h  < u( [imath]\varepsilon[/imath] ) [imath]\rightarrow[/imath]  f(2+h)  4  < [imath]\varepsilon[/imath].
(Can you see the symbols above? I am using a trick to display math symbols on the board, and I cannot tell if it works on your end!)
The trick here is that we make the implicit function that maps epsilon to delta explicit, instead of implicit. Instead of "finding a delta for each epsion", you find that function, fix it, and then show that it works for all epsilon.
This is logically equivalent (really, it is!), but it makes that mapping more explicit, which might help understanding it.
This has the neat property that continuity is about "there is a function at each point", and uniform continuity becomes "there is a function that exists that works for every point".
In that variant, to show a function is continuous you have to produce a modulus of continuity (a function) that, when given the epsilon, produces the delta.
This is done explicitly, instead of explicitly, in the definition.
Ie:
[math]\lim_{x\rightarrow 2} f(x) = 4[/math]
means:
There exists a function u called "the modulus of continuity" such that for all [imath]\varepsilon[/imath] > 0, 0 <  h  < u( [imath]\varepsilon[/imath] ) [imath]\rightarrow[/imath]  f(2+h)  4  < [imath]\varepsilon[/imath].
(Can you see the symbols above? I am using a trick to display math symbols on the board, and I cannot tell if it works on your end!)
The trick here is that we make the implicit function that maps epsilon to delta explicit, instead of implicit. Instead of "finding a delta for each epsion", you find that function, fix it, and then show that it works for all epsilon.
This is logically equivalent (really, it is!), but it makes that mapping more explicit, which might help understanding it.
This has the neat property that continuity is about "there is a function at each point", and uniform continuity becomes "there is a function that exists that works for every point".
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
 squareroot1
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Re: EpsilonDelta proof
Epsilondelta proofs are as follows.
1) Can you get this function within epsilon of the limit?
2) Yep, so long that x is within delta of some value.
2) Here's the delta and the value that work for the function, limit, and that particular epsilon.
1) Neat, now do it for this smaller epsilon!
2) MATHS! HERE'S DELTA!
1) AWESOME! Can you do it for an arbitrary epsilon?!?
2) SURE! Here's a suitable delta in terms of your epsilon!
1) Wowzers! You mean you can make the function arbitrarily close to the limit, provided you use an input suitably close to some value?
2) Isn't that what a limit means?
Epsilondelta proofs are a formality for most users of calculus, anyway.
1) Can you get this function within epsilon of the limit?
2) Yep, so long that x is within delta of some value.
2) Here's the delta and the value that work for the function, limit, and that particular epsilon.
1) Neat, now do it for this smaller epsilon!
2) MATHS! HERE'S DELTA!
1) AWESOME! Can you do it for an arbitrary epsilon?!?
2) SURE! Here's a suitable delta in terms of your epsilon!
1) Wowzers! You mean you can make the function arbitrarily close to the limit, provided you use an input suitably close to some value?
2) Isn't that what a limit means?
Epsilondelta proofs are a formality for most users of calculus, anyway.
Re: EpsilonDelta proof
mikel wrote:DaPwnzlord wrote:I'm going to be taking Calculus I this semester. Last semester I dropped the course early on because I couldn't get the hang of the EpsilonDelta proof. Could you guys give me your best understanding of the proof and how to do it?
Think of it as a badguy good guy game. You are the good guy. The bad guy picks epsilon, you have to produce a delta (or show one exists). Since smaller epsilons are 'harder' (since they encompass all bigger epsilons), expect him to pick really small epsilons.
The most straightforward way to do this is to find an explicit delta in terms of epsilon, but you need to make sure that the delta you get is positive for all (small) epsilon.
So an example.
[math]\lim_{x\rightarrow 2} x^2 = 4[/math]
So, what do we need to show? If some badguy picks a very tiny but positive [imath]\varepsilon[/imath], we need to be able to find a positive [imath]\delta[/imath] so that
[math]x2< \delta \Rightarrow x^2  4 < \varepsilon[/math]
To work with this, we can assume [imath]x2<\delta[/imath], and we want to start with [imath]x^24[/imath] and try to bound it above.
So
[math]x^24=(x2)(x+2) = (x2)(x+2) < \deltax+2[/math]
Ok, this is good, because we now just need to make sure [imath]\deltax+2 < \varepsilon[/imath] and then we are done! But x is close to 2, so x+2 has to be pretty close to 4. Check that as long as we take delta less than 1, then x+2 is positive but less than 5. So now we have [math]x^24<5\delta.[/math] Pick [imath]\delta = \varepsilon/5[/imath] and we are done!
[imath]\delta = min(\varepsilon/5,1)[/imath]
Elvish Pillager wrote:you're basically a daytimemiller: you always come up as guilty to scumdar.
Re: EpsilonDelta proof
mister k wrote:[imath]\delta = min(\varepsilon/5,1)[/imath]
Quite right, I'll edit the original... I was running to a meeting when I finished typing that.
Last edited by mikel on Thu Jan 21, 2010 2:09 pm UTC, edited 1 time in total.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
Re: EpsilonDelta proof
OK, in order to understand the epsilondelta definition of continuity, you need to analyze what a discontinuity in a function is.
Take a graph of f, and suppose that it is discontinuous at x=1, say [imath]f(x)=2x[/imath] for [imath]x\leq 1[/imath], and [imath]f(x)=2x+1[/imath] for [imath]x>1[/imath]. Drawing this on a piece of paper might be a good thing. Looking at the graph, we see that there is a jump at x=1. What do we mean by a jump? What we mean is that [imath]f(x)=2[/imath] at [imath]x=1[/imath], and then a 'little bit' further on it is more than 2. In particular, although to the left of x=1, f(x) is smaller than 2, and right of x=1, f(x) is greater than 3, there was no point close to the discontinuity where f(x) was 2.5, say.
How do we formally write down this sentence though? We say that, if a=1 (the location on the xaxis of the discontinuity), then there is some [imath]\varepsilon[/imath], namely 0.5 (so that [imath]f(a)+\varepsilon=2.5[/imath], such that no matter how close to x=1 you get from the righthand side (i.e., for any [imath]\delta>0[/imath] which we will add to [imath]x[/imath], to get any number a bit more than 1), you never get that [imath]f(a+\delta)=f(a)+\varepsilon=2.5[/imath]. Now, since I could have chosen the [imath]\varepsilon[/imath] to be very slightly smaller, I might as well write this last equality as an inequality, so that you never get that [imath]f(a+\delta)<f(a)+\varepsilon[/imath].
What have we said here? Another way to see it is that, if I draw two horizontal lines across the graph, at y=2.5 and y=1.5 (so that is [imath]f(a)\pm\varepsilon[/imath]), then inside this tube I see some of my function to the left of x=1, but not to the right of x=1. This 'not to the right' business is just saying that, if I go a little bit to the right (i.e., [imath]a+\delta[/imath]), I leave the tube (so that [imath]f(a+\delta)\geq f(a)+\varepsilon[/imath]).
At this point I have lied a bit. I have said 'for any [imath]\delta>0[/imath]', whereas what I meant was 'no matter how close you get to [imath]a[/imath]'. These are different because [imath]\delta[/imath] could be very large, so what we will say is that, for any [imath]\delta>0[/imath], there is some [imath]x[/imath] lying between [imath]a[/imath] and [imath]a+\delta[/imath], such that [imath]f(x)\geq f(a)+\varepsilon[/imath]
Now that we understand what a discontinuity is, to be continuous we have to say that that situation can never arise. For this, we simply negate the statement. The statement above was that f is discontinuous at a if
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } a<x<a+\delta,\text{ with }f(x)f(a)>\varepsilon.[/math]
There are two problems with this. The first is that the function could be decreasing, not increasing! To fix this, which is like saying that the [imath]\varepsilon[/imath] could be negative, we put modulus signs in, to get
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } a<x<a+\delta,\text{ with }f(x)f(a)>\varepsilon.[/math]
This version still needs to be corrected slightly, because it could have been that [imath]f(a)=3[/imath], not [imath]f(a)=2[/imath], so that the jump happened just before [imath]a[/imath], not just after. To get around this problem, we can write
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } a<x<a+\delta,\text{ with }f(x)f(a)>\varepsilon\text{ or there is some }a\delta<x<a,\text{ with }f(x)f(a)>\varepsilon.[/math]
This is slightly clunky, so we again use modulus signs, to get the final form:
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } 0<xa<\delta,\text{ with }f(x)f(a)>\varepsilon.[/math]
This final form is exactly what we want. This says that f is discontinuous at a, so the statement that f is continuous at a is the negation of it, so that
[math]\forall \varepsilon>0,\; \exists \delta>0\text{ such that }\forall 0<xa<\delta,\; f(x)f(a)<\varepsilon.[/math]
This is the definition of continuity.
Edit: I had made a mistake in the previous form as a result of trying to rush...
Take a graph of f, and suppose that it is discontinuous at x=1, say [imath]f(x)=2x[/imath] for [imath]x\leq 1[/imath], and [imath]f(x)=2x+1[/imath] for [imath]x>1[/imath]. Drawing this on a piece of paper might be a good thing. Looking at the graph, we see that there is a jump at x=1. What do we mean by a jump? What we mean is that [imath]f(x)=2[/imath] at [imath]x=1[/imath], and then a 'little bit' further on it is more than 2. In particular, although to the left of x=1, f(x) is smaller than 2, and right of x=1, f(x) is greater than 3, there was no point close to the discontinuity where f(x) was 2.5, say.
How do we formally write down this sentence though? We say that, if a=1 (the location on the xaxis of the discontinuity), then there is some [imath]\varepsilon[/imath], namely 0.5 (so that [imath]f(a)+\varepsilon=2.5[/imath], such that no matter how close to x=1 you get from the righthand side (i.e., for any [imath]\delta>0[/imath] which we will add to [imath]x[/imath], to get any number a bit more than 1), you never get that [imath]f(a+\delta)=f(a)+\varepsilon=2.5[/imath]. Now, since I could have chosen the [imath]\varepsilon[/imath] to be very slightly smaller, I might as well write this last equality as an inequality, so that you never get that [imath]f(a+\delta)<f(a)+\varepsilon[/imath].
What have we said here? Another way to see it is that, if I draw two horizontal lines across the graph, at y=2.5 and y=1.5 (so that is [imath]f(a)\pm\varepsilon[/imath]), then inside this tube I see some of my function to the left of x=1, but not to the right of x=1. This 'not to the right' business is just saying that, if I go a little bit to the right (i.e., [imath]a+\delta[/imath]), I leave the tube (so that [imath]f(a+\delta)\geq f(a)+\varepsilon[/imath]).
At this point I have lied a bit. I have said 'for any [imath]\delta>0[/imath]', whereas what I meant was 'no matter how close you get to [imath]a[/imath]'. These are different because [imath]\delta[/imath] could be very large, so what we will say is that, for any [imath]\delta>0[/imath], there is some [imath]x[/imath] lying between [imath]a[/imath] and [imath]a+\delta[/imath], such that [imath]f(x)\geq f(a)+\varepsilon[/imath]
Now that we understand what a discontinuity is, to be continuous we have to say that that situation can never arise. For this, we simply negate the statement. The statement above was that f is discontinuous at a if
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } a<x<a+\delta,\text{ with }f(x)f(a)>\varepsilon.[/math]
There are two problems with this. The first is that the function could be decreasing, not increasing! To fix this, which is like saying that the [imath]\varepsilon[/imath] could be negative, we put modulus signs in, to get
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } a<x<a+\delta,\text{ with }f(x)f(a)>\varepsilon.[/math]
This version still needs to be corrected slightly, because it could have been that [imath]f(a)=3[/imath], not [imath]f(a)=2[/imath], so that the jump happened just before [imath]a[/imath], not just after. To get around this problem, we can write
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } a<x<a+\delta,\text{ with }f(x)f(a)>\varepsilon\text{ or there is some }a\delta<x<a,\text{ with }f(x)f(a)>\varepsilon.[/math]
This is slightly clunky, so we again use modulus signs, to get the final form:
[math]\exists \varepsilon>0\text{ such that },\; \forall \delta>0\text{ there is some } 0<xa<\delta,\text{ with }f(x)f(a)>\varepsilon.[/math]
This final form is exactly what we want. This says that f is discontinuous at a, so the statement that f is continuous at a is the negation of it, so that
[math]\forall \varepsilon>0,\; \exists \delta>0\text{ such that }\forall 0<xa<\delta,\; f(x)f(a)<\varepsilon.[/math]
This is the definition of continuity.
Edit: I had made a mistake in the previous form as a result of trying to rush...
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