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Homework Help: Evaluate the limit by recognizing that it is a Rieman's sum

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    lim n[itex]\rightarrow[/itex][itex]\infty[/itex] [itex]\frac{16}{n}[/itex]([itex]\sqrt{\frac{16}{n}}[/itex]+[itex]\sqrt{\frac{32}{n}}[/itex]+[itex]\sqrt{\frac{48}{n}}[/itex]+...+[itex]\sqrt{\frac{16n}{n}}[/itex])

    2. Relevant equations



    3. The attempt at a solution

    So by recognizing it is a Rieman's sum, I got to the following conclusion.

    [itex]\frac{64}{n}[/itex][itex]\sum[/itex][itex]\sqrt{\frac{i}{n}}[/itex] (with i=1 under the sigma, and n above it)

    But I don't know if I am right, and even if I am, I don't know how to continue with this! Also, does this replace the lim as n approaches infinity (or rather a better wording would be, does this take care of it?)

    Thanks so much in advance!
     
  2. jcsd
  3. Nov 30, 2011 #2
    Right now, all you have shown is that the limit above is

    [tex]\lim_{n\rightarrow +\infty} \sum_{i=1}^n{\sqrt{\frac{i}{n}}\frac{64}{n}}[/tex]

    You have yet to recognize it as a Riemann sum (and you did not eliminate the limit yet).

    So, can you find a function f such that the above limit equals

    [tex]\lim_{n\rightarrow +\infty} \sum_{i=1}^n{f(x_i)\Delta x_i}[/tex]

    Finding such f and such xi is the clue to this problem.
     
  4. Nov 30, 2011 #3
    I know that in a Rieman sum,

    delta x=(b-a)/n
    so delta x=16/n
    I know that f(xi)=(delta x)(i)
    So f(xi)=(16i)/n

    I am very confused.. I am not even clear on what the steps to solving a problem like this would even be. I think that if I knew such steps I could solve this though.. so it's not the arithmetic.

    Am I trying to show that the original question can be rewritten in terms of


    [tex]\lim_{n\rightarrow +\infty} \sum_{i=1}^n{f(x_i)\Delta x_i}[/tex]

    ?
     
  5. Nov 30, 2011 #4
    Yes, you are trying to rewrite your sum into one of the form

    [tex]\sum_{i=1}^n{f(x_i)\frac{x_i-x_{i-1}}{n}}[/tex]

    It seems obvious that we should try [itex]\frac{x_i-x_{i-1}}{n}=\frac{16}{n}[/itex]. So what are [itex]x_1,...,x_n[/itex] here???
     
  6. Nov 30, 2011 #5
    OK, so after looking at your responses and reading over the book yet again, I discovered I was mistaking that f(x[itex]_{i}[/itex]*)=Δxi, when it is actually just x[itex]_{i}[/itex]* itself that equals Δxi

    I am extremely dense when it comes to the most simple math.. I can see how

    [itex]\frac{x_{i}-x_{i-1}}{n}[/itex]=[itex]\frac{b-a}{n}[/itex], and I already know b to be 16 and a to be 0. So.. x[itex]_{i}[/itex] should be 16, and x[itex]_{i-1}[/itex] should be 0. But I don't quite know how to find x[itex]_{1}[/itex] or x[itex]_{n}[/itex].
     
  7. Nov 30, 2011 #6
    No, [itex]x_1,...,x_n[/itex] are all different!! That's what you do when making a Riemann sum: you divide a certain interval into n (equal) pieces. These pieces are [itex][x_0,x_1], [x_1,x_2],...[/itex].

    You are right that [itex]x_1=0[/itex]. And you also know that [itex]\frac{x_2-x_1}{n}=\frac{16}{n}[/itex]. So, can you now find out what [itex]x_2[/itex] is??

    Can you extend this method to find [itex]x_3[/itex] and so on??
     
  8. Nov 30, 2011 #7
    Yes, I get x[itex]_{2}[/itex]=16, x[itex]_{3}[/itex]=32, x[itex]_{4}[/itex]=48, and so on and so forth. So [itex]\frac{x_{t}}{n}=\frac{16+x_{t-1}}{n}[/itex].

    In otherwords.. I basically just re wrote the original [itex]\frac{x_{i}-x_{i-1}}{n}[/itex]=[itex]\frac{16}{n}[/itex]. I don't really understand what this is supposed to be helping me to find though..
     
  9. Nov 30, 2011 #8
    Oh, I'm sorry, I made a mistake. It turns out that we want

    [tex]x_i-x_{i-1}=\frac{16}{n}[/tex]

    Doing the same thing, we find

    [tex]x_0=0,~x_1=\frac{16}{n},~x_2=\frac{32}{n},...[/tex]

    Why do we do this, well, we wish to find f and [itex]x_i[/itex] (that we just found) to rewrite

    [tex]\sum_{i=1}^n{\sqrt{\frac{16i}{n}}\frac{16}{n}} = \sum_{i=1}^n{f(x_i)(x_i-x_{i-1})}[/tex]

    The right hand side is a Riemann sum. So the limit of that would be an integral. This is why we are doing this. We want to calculate the original sum by calculating the integral.
     
  10. Nov 30, 2011 #9
    Ok that makes a lot of sense! So We have to find a function f in which to plug the value that we found for x[itex]_{i}[/itex], and then multiply it by the value that we found for Δx.
    But I have no idea how to find the function f. I know that somehow I must relate what we found Δx and x[itex]_{i}[/itex] to be to the original question, but how?
     
  11. Nov 30, 2011 #10
    Plug in the values you found for [itex]x_i[/itex] in

    [tex]\sum_{i=1}^n{f(x_i)(x_i-x_{i-1})}[/tex]

    and it will become obvious what f is.
     
  12. Nov 30, 2011 #11
    OK, I did that, and I think I found f(x)=√xi

    OK so I found f, x[itex]_{i}[/itex], and Δx. But now what do I do? I still have no idea..

    Thanks for all of your help by the way, I am just SO frustrated with my prof for rushing over the hardest topic in 15 minutes!
     
  13. Nov 30, 2011 #12
    Ok, so you have found [itex]f(x)=\sqrt{x}[/itex] and [itex]x_i=\frac{16}{n}[/itex].

    Now, I claim that you can write

    [tex]\lim_{n\rightarrow +\infty} \sum_{i=1}^{n} f(x_i)(x_i-x_{i-1}) = \int_a^b f(x)dx[/tex]

    and that integral can be easily calculated. Do you see why you can write it as an integral? It's the very definition of Riemann integrals.

    Ok, so try now to write your original limit as an integral.
     
  14. Nov 30, 2011 #13
    Wait.. would it make sense to just plug what I found for f(x) into the integral with lower limit 0 and upper limit 16? In my mind that would solve the problem.. but then that would mean I found the wrong value for f(x)
     
  15. Nov 30, 2011 #14
    Yes, that is correct. Why do you think that you have found the wrong value??
     
  16. Nov 30, 2011 #15
    OK, thanks! Yes I see how that is equal to the integral. But can you please just go over very quickly how f(x)=sqrt(x)? I get it to be f(x)=sqrt(xi) (not x sub i but actually x times i) which OBVIOUSLY makes no sense and is wrong. It HAS to be wrong, I don't see how I can take the integral of xi.
     
  17. Nov 30, 2011 #16
    Why do you have x TIMES i???

    Isn't it simply because [itex]x_i=\frac{16i}{n}[/itex]??
     
  18. Nov 30, 2011 #17
    Oh I see it now! Thanks so much for sticking with me through all of that, I can understand how frustrating it is to try and teach someone something that you find so simple and they refuse to get it. I am pretty sure I understand the steps now.. I am going to try it again on my own.

    Thanks again!
     
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