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POTD

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  • M MoustafaS

    dnh wrote:

    1 <= ln(x) + 1/x

    That doesn't mean anything, you must prove the whole side : (x-1)/x <= ln x


    About : Islam
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    Jorgen Sigvardsson
    wrote on last edited by
    #20

    Yes it does.

    -- Raaaaaaaaaaaaaaaaaaaaa!

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    • D DavidNohejl

      1<=ln(x)+1/x is equivalent to (x-1)/x <= ln x. I am proving 1<=ln(x)+1/x... And my proof is that for x = 1, ln(x) + 1/x = 1. for x>1, ln(x) + 1/x > 1 because it's raising function, as I proved by showing it's derivation is > 0 on interval (1,e). for x>e, I proved it before.


      "Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe

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      Jorgen Sigvardsson
      wrote on last edited by
      #21

      You are correct. I don't know what he's been smoking. :confused:

      -- Raaaaaaaaaaaaaaaaaaaaa!

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      • M MoustafaS

        This is a pure math question, but its quite good.

        Prove that for all x>=1

        (x-1)/x <= ln x


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        Chris Maunder
        wrote on last edited by
        #22

        Expand as a power series and compare terms.

        cheers, Chris Maunder

        CodeProject.com : C++ MVP

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        • C Chris Maunder

          Expand as a power series and compare terms.

          cheers, Chris Maunder

          CodeProject.com : C++ MVP

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          MoustafaS
          wrote on last edited by
          #23

          Perfect, nice.


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          • C Chris Maunder

            Expand as a power series and compare terms.

            cheers, Chris Maunder

            CodeProject.com : C++ MVP

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            Jorgen Sigvardsson
            wrote on last edited by
            #24

            Why, when it is far simpler to do what dnh did? He basically said that:

            (x-1)/x <= ln x is equivalent to 1<=ln(x)+1/x

            Since we know the properties of ln and ^-1, it's easy to show (by induction) that for all x >= 1, that ln(x) + 1/x >= 1. Series are creepy. :~

            -- Raaaaaaaaaaaaaaaaaaaaa!

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            • J Jorgen Sigvardsson

              Why, when it is far simpler to do what dnh did? He basically said that:

              (x-1)/x <= ln x is equivalent to 1<=ln(x)+1/x

              Since we know the properties of ln and ^-1, it's easy to show (by induction) that for all x >= 1, that ln(x) + 1/x >= 1. Series are creepy. :~

              -- Raaaaaaaaaaaaaaaaaaaaa!

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              Graham Shanks
              wrote on last edited by
              #25

              Can't use mathematical induction. That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work. New POTD: prove that the real numbers are uncountable POTD #2: prove that the rational numbers are countable POTD #3: prove that there are exactly the same number of rational numbers as there are positive integers

              Graham My signature is not black, just a very, very dark blue

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              • G Graham Shanks

                Can't use mathematical induction. That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work. New POTD: prove that the real numbers are uncountable POTD #2: prove that the rational numbers are countable POTD #3: prove that there are exactly the same number of rational numbers as there are positive integers

                Graham My signature is not black, just a very, very dark blue

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                DavidNohejl
                wrote on last edited by
                #26

                Graham Shanks wrote:

                Can't use mathematical induction.

                That's reason why I asked if x was real. But you can easily show that f(x) > f(y) <=> x>y with first derivation of f.


                "Throughout human history, we have been dependent on machines to survive. Fate, it seems, is not without a sense of irony. " - Morpheus "Real men use mspaint for writing code and notepad for designing graphics." - Anna-Jayne Metcalfe

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                • G Graham Shanks

                  Can't use mathematical induction. That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work. New POTD: prove that the real numbers are uncountable POTD #2: prove that the rational numbers are countable POTD #3: prove that there are exactly the same number of rational numbers as there are positive integers

                  Graham My signature is not black, just a very, very dark blue

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                  Jorgen Sigvardsson
                  wrote on last edited by
                  #27

                  Graham Shanks wrote:

                  That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work.

                  Maybe I used the wrong term (didn't know induction was reserved for natural numbers only). If you can show that f(x + dx) > f(x) where dx >, that equation which dnh originally proposed, must be equally valid as any series... right? :~ We know that ln(x+dx) > ln(x) for all x > 1, and we also know that 1/x > 0 for all x > 1. The initial number x = 1 yields ln(1) + 1/1 = 1. I just don't see why you have to bring in fancy pants series to solve this problem. That seems to me like using the sledgehammer just to hit tiny nails...

                  -- Not a substitute for human interaction

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                  • J Jorgen Sigvardsson

                    Graham Shanks wrote:

                    That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work.

                    Maybe I used the wrong term (didn't know induction was reserved for natural numbers only). If you can show that f(x + dx) > f(x) where dx >, that equation which dnh originally proposed, must be equally valid as any series... right? :~ We know that ln(x+dx) > ln(x) for all x > 1, and we also know that 1/x > 0 for all x > 1. The initial number x = 1 yields ln(1) + 1/1 = 1. I just don't see why you have to bring in fancy pants series to solve this problem. That seems to me like using the sledgehammer just to hit tiny nails...

                    -- Not a substitute for human interaction

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                    cp9876
                    wrote on last edited by
                    #28

                    I proved it like dnh - using derivatives - seems perfectly valid to me.


                    Peter "Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."

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                    • G Graham Shanks

                      Can't use mathematical induction. That only works on the natural numbers (one of the steps is to prove that if it works on x then it works for x+1). The real numbers are not countable and therefore induction will not work. New POTD: prove that the real numbers are uncountable POTD #2: prove that the rational numbers are countable POTD #3: prove that there are exactly the same number of rational numbers as there are positive integers

                      Graham My signature is not black, just a very, very dark blue

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                      cp9876
                      wrote on last edited by
                      #29

                      I don't think these are good POTD puzzles, they are straight from a pure maths text. Answers would go like: #1: I think it is Cauchy's famous construction, if they are countable, write them down in order, then construct a new number that is different from the first one in the first decimal place, the second one in the second .... You need a little care to prevent some repeated patterns (9s mainly I htink) you can guarantee to get a number not on the list, QED #2: go around a spiral in Z2 #3: see #2 It is interesting that mathematicians think that there are the same number of rational numbers (that is any number p/q where p and q are integers) as there are integers. There are more real numbers than rational numbers though. You can do funny things with infinities.


                      Peter "Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."

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