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ganeshie8

  • one year ago

show that \[\large \sum\limits_{i=0}^{\infty} \left\lfloor \frac{n+2^i}{2^{i+1}}\right\rfloor = n\] for all positive integers \(n\)

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  1. Empty
    • one year ago
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    Interesting I want to think about this, don't reveal the answer or anything!

  2. ganeshie8
    • one year ago
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    Okay :) I wont say a word because I don't really have a proof for this I have been working on this for a while though, hopelessly..

  3. Empty
    • one year ago
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    Oh in that case tell me everything you know, I thought this was like something you had some clever answer to, but now it sounds like we're gonna need all the help we can get to figure this out.

  4. ganeshie8
    • one year ago
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    I tried to see if this has some connection to the legendre formula for highest exponent of a prime in the prime factorization of \(n!\) http://www.artofproblemsolving.com/wiki/index.php/Legendre's_Formula couldn't use it to my advantage, so i paused on this and trying to use greatest integer function properties and induction

  5. Empty
    • one year ago
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    It seems like it's related to counting in base 2.

  6. Empty
    • one year ago
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    Yeah Legendre's actually sounds like a great idea! I just don't know how to make that work either hmm...

  7. ganeshie8
    • one year ago
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    yeah the successive divisions by 2 clearly has something got to do with base2, but again the numerator expression is not constant.. so its a bit complicated

  8. Empty
    • one year ago
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    I'm trying to see if there's a pattern in here, also where did you find this problem just out of curiosity? https://www.desmos.com/calculator/sydso2ebqk

  9. Empty
    • one year ago
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    We can lower the bound to a finite number. By separating the two terms on top, we can see that these terms will be 0. \[\frac{n}{2^{i+1}} < \frac{1}{2}\] \[\log_2(n)<i\] So maybe this helps us out to write it this way: \[n=\sum_{i=0}^{\lfloor \log_2(n) \rfloor} \left\lfloor \frac{n+2^i}{2^{i+1}} \right\rfloor\]

  10. Empty
    • one year ago
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    I am also looking at this, but it seems to take us nowhere interesting: \[a+b=\sum_{i=0}^{\infty} \left\lfloor \frac{a+b+2^i}{2^{i+1}} \right\rfloor=\sum_{i=0}^{\infty} \left\lfloor \frac{a+2^i}{2^{i+1}} \right\rfloor+\left\lfloor \frac{b+2^i}{2^{i+1}} \right\rfloor\]

  11. ganeshie8
    • one year ago
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    that animation looks interesting, this is from bs grewal's advanced mathematics practice problems

  12. ganeshie8
    • one year ago
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    Adding to that, can we use below consecutive integers property to our advantage Let \[S_n =\sum\limits_{i=0}^{\infty} \left\lfloor \frac{n+2^i}{2^{i+1}}\right\rfloor \] Clearly \(S_1=1\), so it should be sufficient to show that \[S_{n+1}-S_n = 1\]

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