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Math Help - Find x

  1. #1
    Senior Member pankaj's Avatar
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    Find x

    Find the value of :
    \lfloor\frac {x + 1}{2}\rfloor + \lfloor\frac {x + 2}{4}\rfloor + \lfloor\frac {x + 3}{6}\rfloor + \lfloor\frac {x + 4}{8}\rfloor +...........upto \infty
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  2. #2
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    Without going into a detailed analysis, I am sure that this series diverges for all (fixed) values of x.
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  3. #3
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    Hello everyone
    Quote Originally Posted by JG89 View Post
    Without going into a detailed analysis, I am sure that this series diverges for all (fixed) values of x.
    For finite positive x, all terms after a certain point are zero, so it certainly has a finite sum.

    What the answer is, though, I don't know. I'm thinking about it...but I think it may be to do with expressing [an integer] x in binary form.

    Grandad
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  4. #4
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    I just clicked on his Latex source and it shows that he`s talking about the floor function for each term. I thought they were square brackets. How embarassing
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  5. #5
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello everyoneFor finite positive x, all terms after a certain point are zero, so it certainly has a finite sum.

    What the answer is, though, I don't know. I'm thinking about it...but I think it may be to do with expressing [an integer] x in binary form.

    Grandad
    The answer is indeed x,if x is a finite positive number.
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  6. #6
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    Hello pankaj
    Quote Originally Posted by pankaj View Post
    The answer is indeed x,if x is a finite positive number.
    Sorry, I don't really understand what you mean here.

    If f(x) =\lfloor\frac {x + 1}{2}\rfloor + \lfloor\frac {x + 2}{4}\rfloor + \lfloor\frac {x + 3}{6}\rfloor + \lfloor\frac {x + 4}{8}\rfloor + ... then

    f(1) = 1,\, f(2) = 2,\, f(3) = 4,\, f(4) = 5,\, f(5) = 7,\, ...

    and the differences between values of f(x) for consecutive integer values of x don't form any obvious pattern.

    However (using a spreadsheet to do the number-crunching) I have found that for all the values of n I've tried so far (which is up to n = 7), f(2^n) = f(2^{n-1}) + 1. Which is what made me think it might be worth looking at powers of 2 or the binary representation of x.

    Further thoughts: as x increases through consecutive integer values:

    • the value of the first term, \lfloor\frac {x + 1}{2}\rfloor increases by 1 every second value of x; i.e. when x =2, 4, 6, 8, ...
    • the value of the second term, \lfloor\frac {x + 2}{4}\rfloor increases by 1 every 4th value of x; i.e. when x = 2, 6, 10, 14, 18, ...
    • the value of the third term, \lfloor\frac {x + 3}{6}\rfloor increases by 1 every 6th value of x; i.e. when x = 3, 9, 15, 21, ...
    • ... and so on.

    Grandad
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  7. #7
    Senior Member pankaj's Avatar
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    Quote Originally Posted by pankaj View Post
    The answer is indeed x,if x is a finite positive number.
    I am wrong.The post may please be ignored.
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  8. #8
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    Is the answer equal to infinite?
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