Understanding the Riemann Sum - Integral Connection

In summary, the problem is trying to find an approximation to a Riemann integral when approaching infinity. The attempt is to find the sum of a series, but the sum becomes an approximation as the number of rectangles increases. The sum can be approximated by the sum of a function over intervals of length 1/n. Once the intervals are identified, the boundaries of the intervals and the value of the function at the intervals can be found.
  • #1
Rectifier
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The problem
I want to calculate $$\sum^n_{k=1} \frac{4}{1+ \left(\frac{k}{n} \right)^2} \cdot \frac{1}{n}$$ when ##n \rightarrow \infty##

The attempt
## \sum^n_{k=1} \underbrace{f(\epsilon)}_{height} \underbrace{(x_k-x_{k-1})}_{width} \rightarrow \int^b_a f(x) \ dx ##, when ##n \rightarrow \infty## ##n=##amount of rectangles and ##e## is the height of a single rectangle on a specific sub-interval ##k##.

I set ##f(x)=\frac{4}{1+ \left(x \right)^2}## and then for some reason the suggestion in my book is to set ##x_k = \frac{k}{n}## but I don't understand why.

I need help with calculating the end-points for the integral ##a## and ##b## later too.

Pleas help.
 
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  • #2
Rectifier said:
The problem
I want to calculate $$\sum^n_{k=1} \frac{4}{1+ \left(\frac{k}{n} \right)^2} \cdot \frac{1}{n}$$ when ##n \rightarrow \infty##

The attempt
## \sum^n_{k=1} \underbrace{f(\epsilon)}_{height} \underbrace{(x_k-x_{k-1})}_{width} \rightarrow \int^b_a f(x) \ dx ##, when ##n \rightarrow \infty## ##n=##amount of rectangles and ##e## is the height of a single rectangle on a specific sub-interval ##k##.

I set ##f(x)=\frac{4}{1+ \left(x \right)^2}## and then for some reason the suggestion in my book is to set ##x_k = \frac{k}{n}## but I don't understand why.

I need help with calculating the end-points for the integral ##a## and ##b## later too.

Pleas help.

Write out the sum in detail (not using summation notation ##\sum##) and see why the sum is an approximation to a Riemann integral.
 
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  • #3
I can understand why ##x_k=\frac{k}{n}## makes sense, for one wants to have ##\Delta x = \frac{1}{n}##. But how would you plan to deal with the ##k## in ##f(\frac{x}{k})## in your setup?
 
  • #4
Ray Vickson said:
Write out the sum in detail (not using summation notation ##\sum##) and see why the sum is an approximation to a Riemann integral.

##\sum^n_{k=1} \frac{4}{1+ \left(\frac{k}{n} \right)^2} \cdot \frac{1}{n} = \frac{4}{1+ \left(\frac{1}{n} \right)^2} \cdot \frac{1}{n} + \frac{4}{1+ \left(\frac{2}{n} \right)^2} \cdot \frac{1}{n} + ... + \frac{4}{1+ \left(\frac{n}{n} \right)^2} \cdot \frac{1}{n}##

I have no idea how to continue.
 
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  • #5
fresh_42 said:
I can understand why ##x_k=\frac{k}{n}## makes sense, for one wants to have ##\Delta x = \frac{1}{n}##. But how would you plan to deal with the ##k## in ##f(\frac{x}{k})## in your setup?
I am not sure of what you mean by ##f(\frac{x}{k})##
 
  • #6
You wrote ##f(x)=\frac{4}{1+(x)^2}## so your sum is ##\sum_{k=1}^{n}f(\frac{k}{n})\cdot \frac{1}{n}##. I don't see how this helps.
The setting ##x_k=\frac{k}{n}## results in ##\sum_{k=1}^{n}f(x_k)\cdot (x_k-x_{k-1})## which should look familiar to you.
 
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  • #7
Yes that looks familiar indeed but how could I have come up with that by myself without "cheating" and looking on the suggestions-page for this problem? Is there something that I should take into consideration if I were to receive sums like these in the future? Sorry if that is a weird question.
 
  • #8
It's not weird, it's only difficult to answer. I don't know about procedures that automatically lead to results in general. Unfortunately, for otherwise ... Anyway. You normally gain these abilities by experience, trial and error, a lot of reading, a lot of practice, sometimes by pure luck, and some problems might stay difficult. If one way doesn't lead to somewhere, simply try another. It's also a matter of personal attributes. Some are better at recognizing patterns than others, but even this could heavily depend on the patterns in question.

In this special case a possible line of thought could have been:

##S(n) = \sum_{k=1}^n \text{ sth. } \cdot \frac{1}{n}## is a constant and a sum of something added ##n-##times over equidistant points with interval length ##\frac{1}{n}##. The context of Riemann sums might help a lot here. Otherwise one could also conclude: Now if ##n## is increasing, then the intervals become shorter and shorter, so integrals are a natural association. The crucial point here is to regard ##\frac{1}{ n}## as the length of an interval. Once you have this, you can proceed: What are the boundaries of these intervals? What happens with the something at those points? And so on.

Another way might have been to start with the setting ##x(k)=\frac{k}{n}## because this is, what really disturbs. The next question here would be: What is ##\frac{1}{n}## in terms of these ##x(k)##? I think, your only mistake has been by setting ##x=\frac{1}{n}## because it ignores ##k## on which ##x## also depends on.

These are different ways to look at the problem: by a) context (Riemann sums), b) intervals (of length ##\frac{1}{n}##) or c) the function (with the disturbing ##\frac{k}{n}##). You just looked at it one way and got stuck. The only advice that comes to mind and often helps is:
Approach it from another side and only drop information if you're sure you won't need it anymore.
In your case, the fact, that ##x=x(k)=x_k##; but it applies to anywhere else as well: estimations, round ups, the use of unities and many more.
 
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  • #9
Rectifier said:
Yes that looks familiar indeed but how could I have come up with that by myself without "cheating" and looking on the suggestions-page for this problem? Is there something that I should take into consideration if I were to receive sums like these in the future? Sorry if that is a weird question.

Practice and experience. Do lot of problems until the methods become familiar. Besides, I would be very surprised if your textbook did not have already some examples of that type, so you just need to read carefully and fully, and try to reproduce what the book does.
 
  • #10
Ray Vickson said:
Practice and experience. Do lot of problems until the methods become familiar. Besides, I would be very surprised if your textbook did not have already some examples of that type, so you just need to read carefully and fully, and try to reproduce what the book does.
This is my first encounter with Riemann sum like that other were constant.

EDIT:
Wait, I have not had any encounters with Riemann sums at all in this book. I have had an introduction to the concept with a picture with rectangles under the curve with varying width without any examples. The other integrals were step-functions as seen in my last post. Next chapter is about Generalized integrals or something like that.
 
  • #11
Thank you for your help so far everyone it has been helpful in understanding this problem better.

Now to the last issue: how do I calculate the end-points of the integral?

I know that the length of the interval is ##b-a## and ##\frac{b-a}{n}## represents the amount of sub-intervals. Unfortunately that is not making it easier to continue :(
 
  • #12
Rectifier said:
Thank you for your help so far everyone it has been helpful in understanding this problem better.

Now to the last issue: how do I calculate the end-points of the integral?

I know that the length of the interval is ##b-a## and ##\frac{b-a}{n}## represents the amount of sub-intervals. Unfortunately that is not making it easier to continue :(
I haven't done the exercise, so I don't know. But how about my advice? Try it from the other side.
How is ##\int_a^bf (x)dx## defined with Riemann sums? Write the down the limit, compare it with your sum and see where it leads you to. Just have an eye on the indices, i.e. whether it starts at ##0## or ##1##.
 
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  • #13
fresh_42 said:
I haven't done the exercise, so I don't know. But how about my advice? Try it from the other side.
How is ##\int_a^bf (x)dx## defined with Riemann sums? Write the down the limit, compare it with your sum and see where it leads you to. Just have an eye on the indices, i.e. whether it starts at ##0## or ##1##.

How should I write down the limit? Like I did in post #4 and the try to predict what happens when ## n \rightarrow \infty ##?
 
  • #14
No. I meant the integral. How does your book define the integral? ##\int_a^b f(x)dx = \lim_{n\rightarrow\infty} \sum_{k=1}^{n} f(t_k)(x_k-x_{k-1})## with ##t_k \in [x_{k-1},x_k]##? How do ##a## and ##b## come into play here?
 
  • #15
fresh_42 said:
No. I meant the integral. How does your book define the integral? ##\int_a^b f(x)dx = \lim_{n\rightarrow\infty} \sum_{k=1}^{n} f(t_k)(x_k-x_{k-1})## with ##t_k \in [x_{k-1},x_k]##? How do ##a## and ##b## come into play here?
Yes. and also ##a=x_0## and ##b=x_n## but I don't know how to find ##x_0## and ##x_n##?
 
  • #16
C'mon. ##x_k=\frac{k}{n}## and you try to find ##x_0## and ##x_n##?
 
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  • #17
fresh_42 said:
C'mon. ##x_k=\frac{k}{n}## and you try to find ##x_0## and ##x_n##?
0 and 1 but the problem is that I should start my sum at k=1 (##k=1##) or doesn't that matter?
 
  • #18
If you partition an interval into ##n## sections, you will get ##n+1## of boundary points. Think as if you don't have to calculate the sum. Calculate the integral instead. Thus you start at the interval, cut it into pieces and write it down as in #14. Formally it might be some more things to do, because we have ##f(x_k)(x_k-x_{k-1})## instead of ##f(t_k)(x_k-x_{k-1})##. Therefore I asked about your definition. You know, the stuff with upper and lower sums, or infima and suprema or similar. And ##a=0## doesn't mean ##f(a)=0##. You could also make a drawing to see what has to be added.
 
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  • #19
Rectifier said:
This is my first encounter with Riemann sum like that other were constant.

EDIT:
Wait, I have not had any encounters with Riemann sums at all in this book. I have had an introduction to the concept with a picture with rectangles under the curve with varying width without any examples. The other integrals were step-functions as seen in my last post. Next chapter is about Generalized integrals or something like that.

You must be using one very weird textbook. In previous threads you were doing integration by going into the complex plan, doing integrals by partial fractions, etc., and now you are saying the book has not really even defined integrations properly yet (or something like that).

All I can advise is this: if your book is pretty bad, go to the library and take out a better book (or, nowadays, look topics up via Google, etc.) A google search on "riemann integral"; for example, turns up numerous articles, but some of them are quite formal and "abstract", although the article
http://www.geometer.org/mathcircles/riemannint.pdf
is reasonably elementary and covers the basics in a quite readable way. There are others like that as well.
 
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  • #20
Ray Vickson said:
You must be using one very weird textbook. In previous threads you were doing integration by going into the complex plan, doing integrals by partial fractions, etc., and now you are saying the book has not really even defined integrations properly yet (or something like that).

All I can advise is this: if your book is pretty bad, go to the library and take out a better book (or, nowadays, look topics up via Google, etc.) A google search on "riemann integral"; for example, turns up numerous articles, but some of them are quite formal and "abstract", although the article
http://www.geometer.org/mathcircles/riemannint.pdf
is reasonably elementary and covers the basics in a quite readable way. There are others like that as well.

Thank you for your suggestion. Yes, I was doing indefinite integrals (primitive functions) but not integrals with partitions like now.
 
  • #21
fresh_42 said:
If you partition an interval into ##n## sections, you will get ##n+1## of boundary points. Think as if you don't have to calculate the sum. Calculate the integral instead. Thus you start at the interval, cut it into pieces and write it down as in #14. Formally it might be some more things to do, because we have ##f(x_k)(x_k-x_{k-1})## instead of ##f(t_k)(x_k-x_{k-1})##. Therefore I asked about your definition. You know, the stuff with upper and lower sums, or infima and suprema or similar. And ##a=0## doesn't mean ##f(a)=0##. You could also make a drawing to see what has to be added.

Ok, I get it now.

EDIT: I guess the only thing left is to calculate the integral since the sum has to approach the value of the integral (area under the curve) for ##n \rightarrow \infty##, right?

EDIT 2: Looks like I dint get it as I thought before since we have "##f(x_k)(x_k-x_{k-1})## instead of ##f(t_k)(x_k-x_{k-1})##" as you said. I don't know how to deal with that.

EDIT 3: Well, since ## t_k \in [x_k,x_{k-1}]## it isn't prohibited to select a end point of that interval. Hence ##t_k = x_k## and ##f(t_k)(x_k-x_{k-1}) \Rightarrow f(x_k)(x_k-x_{k-1})##. Am I on the right track?
 
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  • #22
Rectifier said:
Ok, I get it now.

EDIT: I guess the only thing left is to calculate the integral since the sum has to approach the value of the integral (area under the curve) for ##n \rightarrow \infty##, right?

EDIT 2: Looks like I dint get it as I thought before since we have "##f(x_k)(x_k-x_{k-1})## instead of ##f(t_k)(x_k-x_{k-1})##" as you said. I don't know how to deal with that.

EDIT 3: Well, since ## t_k \in [x_k,x_{k-1}]## it isn't prohibited to select a end point of that interval. Hence ##t_k = x_k## and ##f(t_k)(x_k-x_{k-1}) \Rightarrow f(x_k)(x_k-x_{k-1})##. Am I on the right track?

For finite sums the value of ##\sum_k f(t_k) (x_k - x_{k-1}) ## may (and usually will) depend on exactly how the ##t_k## are chosen from the intervals ##[x_{k-1},x_k]##. However, for a Riemann-integrable function---for example, for a piecewise-continuous ##f(x)##---you will get the same answer in the limit when all the interval lengths ##x_k - x_{k-1} \to 0## and the number of terms in the sum goes to ##\infty##. In other words, in the limit it will not matter how the ##t_k## are chosen from their intervals.

If you are just a beginner, you can take that at face value, but it is an actual theorem that is proved in some of the more advanced treatments.
 
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What is a Riemann sum?

A Riemann sum is a method used in calculus to approximate the area under a curve by dividing it into smaller, simpler shapes, such as rectangles. It is a foundational concept in the development of the integral.

What is the difference between a left Riemann sum and a right Riemann sum?

A left Riemann sum uses the left endpoint of each subdivision to determine the height of the rectangle, while a right Riemann sum uses the right endpoint. The choice of which type of Riemann sum to use can affect the accuracy of the approximation.

How is a Riemann sum related to the definite integral?

A Riemann sum is a discrete approximation of the definite integral, which represents the exact area under a curve. As the number of subdivisions increases, the Riemann sum approaches the exact value of the definite integral.

What is the significance of the width of the rectangles in a Riemann sum?

The width of the rectangles in a Riemann sum represents the size of the subdivisions used to approximate the area under the curve. The smaller the width, the more accurate the approximation will be.

Can a Riemann sum be used for functions other than polynomials?

Yes, a Riemann sum can be used for any continuous function. However, the more complex the function, the more challenging it may be to accurately approximate the area under the curve using Riemann sums. Other methods, such as the trapezoidal rule or Simpson's rule, may be more effective for certain types of functions.

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