# Epsilon and delta

1. May 16, 2005

### heman

Hi all,,
I came to College and Calculus started with epsilon and delta at beginning...For few weeks i was not even able to get a pich of understanding of it...But somehow i got it...But i find using Epsilon and delta ..solving a question makes it awkard....i want to know does solving the question with the help of these involving... such awkard methods has got practical applications...

I will be thankful for urs replies..

2. May 16, 2005

### inha

Epsilon and delta in what context more precisely? Delta usually denotes change and epsilon a small number. Both have their uses in a vide variety of natural sciences.

3. May 16, 2005

### ahrkron

Staff Emeritus
Most likely he refers to their use in limits.

4. May 16, 2005

### ahrkron

Staff Emeritus
The practical application they have is that they help you understand what rigor means in math.

Don't worry too much about the "practical application". It is too early for that (weird... I feel like an old grandpa saying this).

What is important for you is to develop the skills of thought needed in these methods. Besides helping you understand this particular thing (limits and continuity, and theorems about them later on), the general ability to manipulate these kind of objects in your mind, and to make solid statements with the help of "awkward methods" will be a very valuable one later on.

5. May 16, 2005

### matt grime

Solving eps and delta arguments can be done without thinking.

Suppose we wished to show that x^3 is continuous at z, and let's assume z is strictly positive.

given e>0 we wish to find a d such that |z-y|<d implies |z^3-y^3|<e

|z^3-y^3| = |z-y||z^2+zy+y^2|

well, I can always declare that d<z since if it's true for some positive d it's true for all smaller positive d, and hence one less than z eventually, recalling z is positive.

But then if |z-y|<z it follows 0<z-y<2z

thus I can replace that above epxansion with something bigger, namely |z-y| with d and z^2+zy+y^2 with 3z^2

so the whole thing is less than d3z^2

thus given epsilon, i just pick d such that d<e/3z^2 (and less than |z| too, which I can do), and then

|z-y|<d implies |z^3-y^3|<e as required.

doing it for z=0 and z<0 are easy variations once yu unerstand this case.

6. May 16, 2005

### mathwonk

historically, epsilon and delta had the practical application of ridding analysis of wrong ideas, and opening up to the masses the meaning of correct but previously imprecise ideas like derivatives.

It helps today too, for people who want to understand what the statements mean, and why they are true.

another practical application is making error estimates, since that is rpecisely what epsilon and delta usually are.

if they displaese you, you might try improving on them. i.e. what would you offer as a precise definition of the statement that a certain sequence of non zero numbers nonetheless "approaches" zero?

i.e. how would you go about explaining how an infinite sequence of numbers can be used to determine a different number, their "limiting value"?

Last edited: May 16, 2005
7. May 17, 2005

### heman

Thnks very much for urs replies...

Matt thanks for urs this method but i am such a big fool i am not able to understand this too well ...Can u illustrate more these lines...

"well, I can always declare that d<z since if it's true for some positive d it's true for all smaller positive d, and hence one less than z eventually, recalling z is positive.

But then if |z-y|<z it follows 0<z-y<2z""

Also is it feasible in other questions....Can u give me 1 more example which does it in similar way...

Last edited: May 17, 2005
8. May 17, 2005

### matt grime

Suppose something is true for all x in the interval [y-1,y+1], then it must be true for x in any subinterval of [y-1,y+1], mustn't it? That's all I'm saying.

So, in this case I can say that since I want to find a d and consider the interval [z-d,z+d] can assume this interval is inside the interval [0,2z] by assuming d is less than z.

I need to get some upper bounds, and I am free to do whatever I wish to create this upper bound.

Do it for the special case of z=1

Given e i need to find a d such that |y-1|<d implies |1-y^3|<e.

First I may assume d<1, do you get why that is ok to assume?

Then |y-1|<1 is the same as -1<y-1<1, 0<y<2

so if |y-1|<d<1 implies 0<y<2

so that |1-y^3| = |1-y||1+y+y^2| < d(1+2+4)=6d

so, as long as d is less than one and less than e/6 which is simultaneously possible it follows that

|1-y^3|<6d < e

ok?

9. May 17, 2005

### heman

I got it well Matt...This was very well explained..Thanks very much.