- #1
Determination of limit in Riemann sums.
- Thread starter alingy1
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In summary, a Riemann sum is a method used in calculus to approximate the area under a curve by dividing it into smaller rectangles and summing their areas. The limit in a Riemann sum is determined by taking the limit as the width of the rectangles approaches zero. There are two types of Riemann sums, left and right, which use different endpoints to calculate the approximate area under a curve. Riemann sums can be used to approximate non-rectangular shapes by dividing the curve into smaller sections. They are also closely related to integrals, as the Riemann sum becomes more accurate as the width of the rectangles approaches zero, approaching the value of the integral.
- #2
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Have you tried summation by parts? Try setting ##a_i = i##, ##b_i = e^{-2i/n}##, and ##B_i = \sum_{j=1}^i b_j##. Then your sum becomes
$$\begin{align}
\frac{1}{n^2}\sum_{i=1}^n a_i b_i
&= \frac{1}{n^2} \left(a_n b_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) \end{align}$$
which should be easier to work with since ##a_{i+1} - a_i = 1## and ##B_i## is a partial sum of a geometric series.
$$\begin{align}
\frac{1}{n^2}\sum_{i=1}^n a_i b_i
&= \frac{1}{n^2} \left(a_n b_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) \end{align}$$
which should be easier to work with since ##a_{i+1} - a_i = 1## and ##B_i## is a partial sum of a geometric series.
- #3
HallsofIvy
Science Advisor
Homework Helper
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Since you posted another problem with "Riemann sums", yes, that is probably what you are expected to do. I doubt there is any "other way" to do this.
Now, do you know how, without using "Riemann sums", to find [itex]\int_0^1 xe^{-2x} dx[/tex]?
(Hint- let u= -2x.)
Now, do you know how, without using "Riemann sums", to find [itex]\int_0^1 xe^{-2x} dx[/tex]?
(Hint- let u= -2x.)
- #4
alingy1
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Jbunnii, i followed the site. Apparently that method is similar to integration by parts and it looks quite complicated!
I think this clarified that i must aimply use the integral which is quite easy to do with integration by parts!
Thanks for the help!
I think this clarified that i must aimply use the integral which is quite easy to do with integration by parts!
Thanks for the help!
- #5
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[edited to fix a few typos]
The summation by parts approach also works, though. It has the advantage of being more elementary (you don't need to bring in the machinery of Riemann integration) but it does require a bit more work. Here's a sketch. As I wrote above, we set ##a_i = i##, ##b_i = e^{-2i/n}##, and ##B_i = \sum_{j=1}^{i} b_j##. Then we have to evaluate the limit of this sum:
$$\begin{align}
\frac{1}{n^2}\sum_{i=1}^n a_i b_i
&= \frac{1}{n^2} \left(a_n b_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) \end{align}$$
Note that ##a_{i+1} - a_i = 1## and
$$B_i = \sum_{j=1}^{i} e^{-2j/n} = \sum_{j=1}^{i} (e^{-2/n})^j = e^{-2/n}\sum_{j=0}^{i-1}(e^{-2/n})^j = e^{-2/n}\frac{1 - e^{-2i/n}}{1 - e^{-2/n}}$$
So our sum reduces to
$$\frac{1}{n^2} \left(n e^{-2} - \frac{e^{-2/n}}{1 - e^{-2/n}} \sum_{i=1}^{n-1} (1 - e^{-2i/n})\right)$$
Now you need to simplify further and take the limit as ##n \rightarrow \infty##.
I agree, the integral/Riemann sum method should be the easiest way to solve this, and most likely it's what the author of the question intended.alingy1 said:
The summation by parts approach also works, though. It has the advantage of being more elementary (you don't need to bring in the machinery of Riemann integration) but it does require a bit more work. Here's a sketch. As I wrote above, we set ##a_i = i##, ##b_i = e^{-2i/n}##, and ##B_i = \sum_{j=1}^{i} b_j##. Then we have to evaluate the limit of this sum:
$$\begin{align}
\frac{1}{n^2}\sum_{i=1}^n a_i b_i
&= \frac{1}{n^2} \left(a_n b_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) \end{align}$$
Note that ##a_{i+1} - a_i = 1## and
$$B_i = \sum_{j=1}^{i} e^{-2j/n} = \sum_{j=1}^{i} (e^{-2/n})^j = e^{-2/n}\sum_{j=0}^{i-1}(e^{-2/n})^j = e^{-2/n}\frac{1 - e^{-2i/n}}{1 - e^{-2/n}}$$
So our sum reduces to
$$\frac{1}{n^2} \left(n e^{-2} - \frac{e^{-2/n}}{1 - e^{-2/n}} \sum_{i=1}^{n-1} (1 - e^{-2i/n})\right)$$
Now you need to simplify further and take the limit as ##n \rightarrow \infty##.
- #6
Saitama
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This too can be handled without the definite integrals. First notice that:
$$\sum_{i=1}^{n} x^i=x\frac{x^n-1}{x-1}$$
(You should be able to prove the above as it is a simple geometric progression).
Differentiate both the sides with respect to ##x## to obtain:
$$\sum_{i=1}^n i\,x^{i-1}=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$
$$\Rightarrow \sum_{i=1}^n i\,x^{i}=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$$
Replace ##x## with ##e^{-2/n}## and divide by ##n^2## on both the sides to obtain:
$$\sum_{i=1}^n \frac{ie^{-2i/n}}{n^2}=\frac{ne^{-2-4/n}-(n+1)e^{-2-2/n}+e^{-2/n}}{(e^{-2/n}-1)^2n^2}=\frac{n+e^{2/n+2}-e^{2/n}(n+1)}{e^2(e^{2/n}-1)^2n^2}$$
$$\Rightarrow \sum_{i=1}^n \frac{ie^{-2i/n}}{n^2}=\frac{-n(e^{2/n}-1)+e^{2/n}(e^2-1)}{e^2(e^{2/n}-1)^2n^2}=\frac{-2\left(\frac{e^{2/n}-1}{2/n}\right)+e^{2/n}(e^2-1)}{4e^2\left(\frac{e^{2/n}-1}{2/n}\right)^2}$$
Notice that as ##n\rightarrow \infty##, ##e^{2/n} \rightarrow 1## and ##\frac{e^{2/n}-1}{2/n} \rightarrow 1##, hence,
$$\lim_{n\rightarrow \infty}\frac{-2\left(\frac{e^{2/n}-1}{2/n}\right)+e^{2/n}(e^2-1)}{4e^2\left(\frac{e^{2/n}-1}{2/n}\right)^2}=\frac{-2+e^2-1}{4e^2}=\boxed{\dfrac{e^2-3}{4e^2}}$$
Moral: It is better to use definite integrals.
$$\sum_{i=1}^{n} x^i=x\frac{x^n-1}{x-1}$$
(You should be able to prove the above as it is a simple geometric progression).
Differentiate both the sides with respect to ##x## to obtain:
$$\sum_{i=1}^n i\,x^{i-1}=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$
$$\Rightarrow \sum_{i=1}^n i\,x^{i}=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$$
Replace ##x## with ##e^{-2/n}## and divide by ##n^2## on both the sides to obtain:
$$\sum_{i=1}^n \frac{ie^{-2i/n}}{n^2}=\frac{ne^{-2-4/n}-(n+1)e^{-2-2/n}+e^{-2/n}}{(e^{-2/n}-1)^2n^2}=\frac{n+e^{2/n+2}-e^{2/n}(n+1)}{e^2(e^{2/n}-1)^2n^2}$$
$$\Rightarrow \sum_{i=1}^n \frac{ie^{-2i/n}}{n^2}=\frac{-n(e^{2/n}-1)+e^{2/n}(e^2-1)}{e^2(e^{2/n}-1)^2n^2}=\frac{-2\left(\frac{e^{2/n}-1}{2/n}\right)+e^{2/n}(e^2-1)}{4e^2\left(\frac{e^{2/n}-1}{2/n}\right)^2}$$
Notice that as ##n\rightarrow \infty##, ##e^{2/n} \rightarrow 1## and ##\frac{e^{2/n}-1}{2/n} \rightarrow 1##, hence,
$$\lim_{n\rightarrow \infty}\frac{-2\left(\frac{e^{2/n}-1}{2/n}\right)+e^{2/n}(e^2-1)}{4e^2\left(\frac{e^{2/n}-1}{2/n}\right)^2}=\frac{-2+e^2-1}{4e^2}=\boxed{\dfrac{e^2-3}{4e^2}}$$
Moral: It is better to use definite integrals.
- #7
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There's a key error here: That ##a_nb_n## should be ##a_nB_n##. This should bejbunniii said:
[tex]\frac{1}{n^2}\sum_{i=1}^n a_i b_i
= \frac{1}{n^2} \left(a_n B_n - \sum_{i=1}^{n-1}B_i (a_{i+1}-a_i)\right) [/tex]
- #8
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- 257
Good catch, thanks!D H said:
Related to Determination of limit in Riemann sums.
1. 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 rectangles and summing their areas. It is an important concept in the study of limits and integrals.
2. How do you determine the limit in a Riemann sum?
The limit in a Riemann sum is determined by taking the limit as the width of the rectangles approaches zero. This is known as the limit of a sum and can be calculated using techniques such as the squeeze theorem or the definition of a limit.
3. What is the difference between a left Riemann sum and a right Riemann sum?
A left Riemann sum uses the left endpoints of the rectangles to calculate the approximate area under a curve, while a right Riemann sum uses the right endpoints. The difference between the two sums can give an indication of how closely the sum approximates the actual area under the curve.
4. Can Riemann sums be used to approximate non-rectangular shapes?
Yes, Riemann sums can be used to approximate the area under any curve, even if it is not a rectangular shape. This can be done by dividing the curve into smaller sections and approximating the area of each section using rectangles.
5. How do Riemann sums relate to integrals?
Riemann sums are used to approximate the area under a curve, which is the same concept as finding the definite integral of a function. As the width of the rectangles approaches zero, the Riemann sum becomes more accurate and approaches the value of the integral.
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