Solution 75: Integrating Limit, Arithmetic Mean, and Trig

  • Published on Jan 17, 2019
  • Hmm, we have n approaching infinity and a factor of 1/n residing inside arithmetic means. Can you guess what this limit is equal to?
    Congratulations to Bamdad Shamaei, Parth Pawar, PRAKHAR AGARWAL, Kwekinator117, 123 gogo, Nicholas Patel, Jaleb, Hiren Bavaskar, Arun Bharadwaj, and Isaac YIU Math Studio for successfully solving this math challenge question! Bamdad Shamaei was the first person to solve the question.
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Comments • 24

  • Sitanshu Chaudhary
    Sitanshu Chaudhary 2 months ago +1

    But i have doubt what is the limit n tends infinite sin(2ln(n)) is converge

  • aaronlabuka
    aaronlabuka 6 months ago

    waw, impressing

  • trey hanckok
    trey hanckok 8 months ago

    I tried to integrate by parts however my solution included ln0 which is und. Any thoughts?

    • Aaron He
      Aaron He 6 months ago

      Take the limit and ln0 is -infinity.

  • Jeffrey Cloete
    Jeffrey Cloete 8 months ago +2

    Sir..whomever you are a brilliant problem poser..and have an ingenious faculty for maths. Thank you for sharing your skill on USclip! !

    • LetsSolveMathProblems
      LetsSolveMathProblems  7 months ago

      Thank you for your heartwarming compliment. I'm glad you find my videos enjoyable and educational! =)

  • 민뚜TV
    민뚜TV 8 months ago


  • ツGMD Novacraft
    ツGMD Novacraft 8 months ago +3

    like if u dont understand nothing but still liking this vids

  • Rot
    Rot 8 months ago

    Where can I learn about this

  • Mike Fernandez
    Mike Fernandez 8 months ago

    Love your videos, I always learn something new!

  • Janda125
    Janda125 8 months ago +8

    You put both sin(ln n) and sin(ln i) = A, same with the cosines, you put cos(ln n) and cos(ln i) as B

    • Janda125
      Janda125 8 months ago +3

      LetsSolveMathProblems No worries, it was a great video, no need to apologize

    • LetsSolveMathProblems
      LetsSolveMathProblems  8 months ago +3

      Oops! You are correct. I apologize for the error.

    • Janda125
      Janda125 8 months ago +1

      I know it wasn’t the exact expretion, I just wanted to point out to you which ln n and ln i belonged to which cos and sin. I think you ment ln n = A and ln i = B, right?

  • Furious Fthefics
    Furious Fthefics 8 months ago +1

    I can do it easily but how I can post my solution

    • Felipe Lorenzzon
      Felipe Lorenzzon 8 months ago +1

      I think you can post it elsewhere as a picture or with latex and put the link to it in your comment. But I would recommend you to give at least a brief description of your solution and also your final answer

    • LetsSolveMathProblems
      LetsSolveMathProblems  8 months ago

      The solution to this weekly math challenge had to be submitted on the "Problem 75" video by yesterday (January 15, 2019) at the latest, so it is too late to turn in any solution to this specific problem. If you are interested, here is a link to this week's challenge problem:

  • Ben Burdick
    Ben Burdick 8 months ago +1

    That was a really cool solution, but It's unclear to me how we determined the limits of integration. Why was it 0 to 1?

    • LetsSolveMathProblems
      LetsSolveMathProblems  8 months ago +3

      You are right (but make sure you also have 5i/n inside the function, along with the factor 5/n outside the summation). Here is the actual limit definition of the definite integral from 0 to 5 of sin(ln(x))dx using right Riemann Sum:
      Limit as n -> infinity of (5/n) * Sum from i=1 to n of ( sin(ln( 5i/n )) ).

    • Ben Burdick
      Ben Burdick 8 months ago +2

      @LetsSolveMathProblems Ah, ok this makes sense. So if I wanted to integrate from 0 to 5, the summation would have been in the form 5/n?

    • LetsSolveMathProblems
      LetsSolveMathProblems  8 months ago +3

      Excellent question. To see why the limits of integration were chosen as in the video, I will give two examples with different limits of integration.
      1. Integral from 0 to 2: In this case, the width of each rectangle is 2/n, NOT 1/n, (because we are dividing an interval of length 2 into n subintervals) and, more importantly, we should evaluate our function at 2i/n, NOT i/n.
      2. Integral from 1 to 2: In this case, we should evaluate of our function at 1+i/n (that is, 1+1/n, 1+2/n, ..., 1+n/n = 2), NOT i/n.
      In our example, our function is being evaluated at 0+i/n, so the bounds should be from 0 to 1.

  • James Wilson
    James Wilson 8 months ago

    Sweet problem! And very good solution Bamdad Shamaei!

  • adandap
    adandap 8 months ago

    I failed on this one last week, though got it out when I had another look at it this morning. But I was quite pleased with one aspect of my failed attempt, which at least let let me bound the answer. I took a complex combination c_n = a_n + i b_n. Then it's not hard to see that the average of c_n = 1/n sum(j=1,n) exp(i ln(j) ), which has modulus

  • Andrés Robles
    Andrés Robles 8 months ago +2

    Tus retos son únicos 👏.