Equal partition Integral problem

In summary, the book claims that for any integrable function there is a partition of the domain into n intervals such that the integral over the whole domain is less than or equal to some constant. However, this is only true for continuous functions. For discontinuous functions, there is no guarantee that the integral will be less than or equal to a certain constant. Additionally, the book claims that for any rational number, if the fraction is raised to a power then the result is 1. However, this is not always the case.
  • #1
subsonicman
21
0
This is spivak's calculus 2nd edition #12b.

The question in part a defines Pn as a partition of [a,b] into n equal intervals.

The question in part b states: Find an integrable function f on [0,1] and an ε > 0 such that U(f,Pn)-∫0 →1 (f) > ε for all n.

I'm made no progress on this at all.
Part (a) asks to prove that there is no such ε if f is continuous, and I proved that using uniform continuity. I've been trying to create a function that has little jumps at all the rational numbers and none anywhere else to make a counterexample, but that gets me nowhere.

Also, I learned that apparently there are different types of integrals, so I think that this book uses the Riemann Stieltjes Integral.
 
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  • #2
What about taking f to be 0 if x is irrational, 1 if x is rational? That should work.
 
  • #3
That function f isn't integrable. Since in every interval there is some x so f(x)=1 and some x so f(x)=0, the upper sum is 1 and the lower sum is 0.
 
  • #4
Can you modify what Halls wrote down in any way?
 
  • #5
WannabeNewton said:
Can you modify what Halls wrote down in any way?

A small hint:

 
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  • #6
WannabeNewton said:
Can you modify what Halls wrote down in any way?

There are two things I can think of. One is an example they use a lot in the book which is 0 for irrationals and 1/q for a rational with p/q in lowest form. But the upper bound can be arbitrarily close to 0 for equal partition.

The second is let the value be 1 for all rationals of the form 1/m and 0 otherwise. That is definitely integrable but it's hard to tell if equal partitions could show it's integrable.

Also, I must be missing something, but how is popcorn a hint?
 
  • #7
subsonicman said:
There are two things I can think of. One is an example they use a lot in the book which is 0 for irrationals and 1/q for a rational with p/q in lowest form. But the upper bound can be arbitrarily close to 0 for equal partition.

The second is let the value be 1 for all rationals of the form 1/m and 0 otherwise. That is definitely integrable but it's hard to tell if equal partitions could show it's integrable.

Clearly ##\int_0^1 f = 0## in both cases. Can you calculate ##U(f,P_n)## for both functions?

Also, I must be missing something, but how is popcorn a hint?

http://en.wikipedia.org/wiki/Popcorn_function
 
  • #8
It's really hard to tell. I thought it was obvious earlier but I was only thinking of the values of the function at the end points.
 
  • #9
subsonicman said:
That function f isn't integrable. Since in every interval there is some x so f(x)=1 and some x so f(x)=0, the upper sum is 1 and the lower sum is 0.
Oh, right. You did say "Riemann Stieltjes Integral".
 
  • #10
@Micromass

I'm fairly sure for the first function ##\inf(U(f,P_n))=0##.

A non-rigorous proof is as follows:
Let's say we want to find ##n## such that ##U(f,P_n)<\frac{1}{m}## Note that there's only a finite number of rationals ##x## such that ##f(x) \ge \frac{1}{m}##. Each of these can only be in a maximum of two intervals in the partition ##P_n##. Every other interval will have a max value of ##\frac{1}{m+1}## or less. Thus if you make the intervals small enough, eventually all the intervals with max values ##\frac{1}{m+1}## or less will "outweigh" the small number of intervals with max values greater than ##\frac{1}{m+1}##.

You can do a similar kind of proof for the second function to show ##\inf(U(f,P_n))=0##.

While of course it's not rigorous (I don't want to go through the trouble of working with these fractions), is there any problem with my argument?

Also, very strangely, in the appendix to the chapter, the book says this:

"The moral of this tale is that anything which looks like a good approximation to an integral really is, provided that all the lengths ##t_i - t_{i-1}## of the intervals in the partition are small enough. We've proved that this only for continuous ##f##, but it actually holds for any integrable ##f##."

This is said after proving that if ##f## is continuous on ##[a,b]##, then for every ##\epsilon > 0## there is a ##\delta > 0## such that for a partition ##P = \{t_0,\cdots,t_n\}## of ##[a,b]##, with each piece of the partition with a width smaller than ##\delta## then ##\mid \sum_{i=1}^n f(x_i)(t_i-t_{i-1}) - \int_a^b f(x)\,dx. \mid < \epsilon ##.How is possible for the problem in my original post and this statement to be true?
 
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FAQ: Equal partition Integral problem

What is the "Equal Partition Integral problem"?

The Equal Partition Integral problem is a mathematical problem that involves finding equal integral partitions of a given number. This means finding two or more integers that add up to the given number.

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What is an example of the "Equal Partition Integral problem"?

An example of the Equal Partition Integral problem would be finding two integers that add up to 10. The possible solutions for this problem are 5 and 5, 3 and 7, or 2 and 8.

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What are some challenges in solving the "Equal Partition Integral problem"?

The Equal Partition Integral problem can be challenging because it involves finding all possible combinations of integers that add up to a given number. It is also difficult to determine the most efficient approach for solving the problem, as it depends on the given number and the available resources. Additionally, the problem can become more complex when considering larger or negative numbers.

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