Euler's summation formula proof

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The following proof is from Apostol's book:
enter image description here



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Questions:



  1. On the first line of the proof, he uses '' just as brackets or do they have other meaning like $[x]$ being the floor function?


  2. right before equation (6), why does the summation from $m+1$ up to $k$ become $kf(k)-mf(m)$?


  3. at equation (6) when he substitutes back $x,y$ i'm not sure why are the two integrals equal?




number-theory elementary-number-theory analytic-number-theory




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  • On 2), see en.wikipedia.org/wiki/Telescoping_series.
    – joriki
    2 days ago














up vote
0
down vote

favorite












The following proof is from Apostol's book:
enter image description here



enter image description here



Questions:



  1. On the first line of the proof, he uses '' just as brackets or do they have other meaning like $[x]$ being the floor function?


  2. right before equation (6), why does the summation from $m+1$ up to $k$ become $kf(k)-mf(m)$?


  3. at equation (6) when he substitutes back $x,y$ i'm not sure why are the two integrals equal?




number-theory elementary-number-theory analytic-number-theory




share|cite|improve this question





















  • On 2), see en.wikipedia.org/wiki/Telescoping_series.
    – joriki
    2 days ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The following proof is from Apostol's book:
enter image description here



enter image description here



Questions:



  1. On the first line of the proof, he uses '' just as brackets or do they have other meaning like $[x]$ being the floor function?


  2. right before equation (6), why does the summation from $m+1$ up to $k$ become $kf(k)-mf(m)$?


  3. at equation (6) when he substitutes back $x,y$ i'm not sure why are the two integrals equal?




number-theory elementary-number-theory analytic-number-theory




share|cite|improve this question













The following proof is from Apostol's book:
enter image description here



enter image description here



Questions:



  1. On the first line of the proof, he uses '' just as brackets or do they have other meaning like $[x]$ being the floor function?


  2. right before equation (6), why does the summation from $m+1$ up to $k$ become $kf(k)-mf(m)$?


  3. at equation (6) when he substitutes back $x,y$ i'm not sure why are the two integrals equal?





number-theory elementary-number-theory analytic-number-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









Spasoje Durovic

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  • On 2), see en.wikipedia.org/wiki/Telescoping_series.
    – joriki
    2 days ago
















  • On 2), see en.wikipedia.org/wiki/Telescoping_series.
    – joriki
    2 days ago















On 2), see en.wikipedia.org/wiki/Telescoping_series.
– joriki
2 days ago




On 2), see en.wikipedia.org/wiki/Telescoping_series.
– joriki
2 days ago










1 Answer
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  1. They are simply brackets, no specific meanings.

  2. Try to write all of them. Then the terms cancelled like this:
    $$
    3f(3) - 2f(2) + 2f(2) - 1f(1) = 3f(3) - 1f(1).
    $$

  3. Since
    $$
    int_k^x lfloor t rfloor f'(t) mathrm d t = k int_k^x f'(t) mathrm d t = f(k) - f(x).
    $$
    Same for the other term.





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    1 Answer
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    1 Answer
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    up vote
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    1. They are simply brackets, no specific meanings.

    2. Try to write all of them. Then the terms cancelled like this:
      $$
      3f(3) - 2f(2) + 2f(2) - 1f(1) = 3f(3) - 1f(1).
      $$

    3. Since
      $$
      int_k^x lfloor t rfloor f'(t) mathrm d t = k int_k^x f'(t) mathrm d t = f(k) - f(x).
      $$
      Same for the other term.





    share|cite|improve this answer

























      up vote
      2
      down vote













      1. They are simply brackets, no specific meanings.

      2. Try to write all of them. Then the terms cancelled like this:
        $$
        3f(3) - 2f(2) + 2f(2) - 1f(1) = 3f(3) - 1f(1).
        $$

      3. Since
        $$
        int_k^x lfloor t rfloor f'(t) mathrm d t = k int_k^x f'(t) mathrm d t = f(k) - f(x).
        $$
        Same for the other term.





      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        1. They are simply brackets, no specific meanings.

        2. Try to write all of them. Then the terms cancelled like this:
          $$
          3f(3) - 2f(2) + 2f(2) - 1f(1) = 3f(3) - 1f(1).
          $$

        3. Since
          $$
          int_k^x lfloor t rfloor f'(t) mathrm d t = k int_k^x f'(t) mathrm d t = f(k) - f(x).
          $$
          Same for the other term.





        share|cite|improve this answer













        1. They are simply brackets, no specific meanings.

        2. Try to write all of them. Then the terms cancelled like this:
          $$
          3f(3) - 2f(2) + 2f(2) - 1f(1) = 3f(3) - 1f(1).
          $$

        3. Since
          $$
          int_k^x lfloor t rfloor f'(t) mathrm d t = k int_k^x f'(t) mathrm d t = f(k) - f(x).
          $$
          Same for the other term.






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        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









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