What is the infimum of b - a for b>a?

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In summary, this person is trying to find the darboux sum for a function where f(x) = 1 if x = 0 and f(x) = 0 otherwise. He got the lower darboux sum and then when he tried to do the upper darboux sum, he got \Sigma(xj+1 - xj)sup f. Clearly, the inf is 0, but he needs to prove it.
  • #1
Gavins
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Homework Statement


If b>a, find the infimum of b - a for arbritrary b,a.


Homework Equations


-


The Attempt at a Solution


So this is for a darboux sum but I'm stuck on proving that 0 is the inf. It seems really easy but I can't seem to think straight.

First thing is 0 is a lower bound of b - a.
b - a > 0
b > a

Next is to prove it's the greatest lower bound. I tried to do this by supposing x > 0 is the greatest lower bound and find some contradiction but I can't figure it out.
 
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  • #2


Ok, go ahead. Suppose x>0 is a lower bound. Pick b=x/2 and a=0. Is x a lower bound for b-a?
 
  • #3


Does that not only prove it then when a = 0? I guess I might as well say what I'm doing. Basically, I'm trying to get the darboux sum for a function where f(x) = 1 if x = 0 and f(x) = 0 otherwise. I got the lower darboux sum and then when I try to do the upper darboux sum, I get [tex]\Sigma[/tex](xj+1 - xj)sup f. So if I expand this sum, I get 0 for every interval except the interval that contains 0 ie [xk,xk+1]. The upper darboux sum is xk+1 - xk. Clearly, the inf is 0 but I need to prove this.
 
  • #4


Gavins said:
Does that not only prove it then when a = 0? I guess I might as well say what I'm doing. Basically, I'm trying to get the darboux sum for a function where f(x) = 1 if x = 0 and f(x) = 0 otherwise. I got the lower darboux sum and then when I try to do the upper darboux sum, I get [tex]\Sigma[/tex](xj+1 - xj)sup f. So if I expand this sum, I get 0 for every interval except the interval that contains 0 ie [xk,xk+1]. The upper darboux sum is xk+1 - xk. Clearly, the inf is 0 but I need to prove this.

Well, no. The assumption that was that x>0 was a lower bound of b-a for ALL b>a. If it's not a lower bound for SOME b>a, then x>0 is NOT a lower bound. But you are thinking about this way too generally. Of course you can make the upper sum as small as you want by picking x_(j+1)-x_j as small as you want. You might make the interval containing 0 be [-1/n,1/n]. So sure, you can make that as close to zero as you want by picking n large enough.
 
  • #5


Thanks for that. Got it all sorted out now.
 

1. What is an infimum in mathematics?

An infimum, also known as greatest lower bound, is the greatest number in a set that is less than or equal to all other numbers in the set. It is denoted by inf(S) or infimum of S.

2. How is infimum different from minimum in mathematics?

The minimum is the smallest number in a set, while the infimum is the greatest number that is less than or equal to all other numbers in the set. In other words, the minimum is a part of the set, while the infimum may or may not be a part of the set.

3. What is a Darboux sum in calculus?

A Darboux sum is a way of approximating the area under a curve using rectangles. It is calculated by dividing the interval into smaller subintervals and taking the sum of the products of the subinterval length and the function value at a specific point in each subinterval.

4. How is a Darboux sum different from a Riemann sum?

Both Darboux and Riemann sums are ways of approximating the area under a curve, but they differ in the choice of the sample points used to calculate the rectangle heights. In Darboux sums, the sample points are chosen at either the left or right endpoint of each subinterval, while in Riemann sums, the sample points can be chosen at any point within the subinterval.

5. What is the significance of infimum and Darboux sums in calculus?

Infimum and Darboux sums are important concepts in calculus because they are used to define the definite integral. The definite integral is a fundamental tool in calculus, used to calculate the area under a curve and solve various problems related to motion, optimization, and more.

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