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experimentX Group Title

Math.

  • 2 years ago
  • 2 years ago

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  1. experimentX Group Title
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    it is given to prove that x/(1+x) < ln(1+x) < x in the interval [1,1+t] t>0

    • 2 years ago
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    |dw:1350718620013:dw|

    • 2 years ago
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    |dw:1350718731005:dw|

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    |dw:1350718795042:dw|

    • 2 years ago
  5. Callisto Group Title
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    But then, you still haven't considered lnx...

    • 2 years ago
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    why do you need to prove ... ln(2) > 1/2 ?

    • 2 years ago
  7. Callisto Group Title
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    It's not prove.. but.. we have to consider lnx..

    • 2 years ago
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    why??

    • 2 years ago
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    By considering ln x in [1; 1 + t] (t > 0), ln x in [1; 1 + t] (t > 0), minimum bound : ln(x) = 1 => x = e

    • 2 years ago
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    x/(1+x) < ln(1+x) < x ... perhaps we can use the same argument as we used above.

    • 2 years ago
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    \[ (x+1)^2 > 1+x > 1 \text{ for all x > 0} \\ {1 \over (1+x)^2} < {1 \over 1 +x} < 1 \text{ for all x > 0} \\ \int_0^x {1 \over (1+x)^2} \;dx < \int_0^x {1 \over 1 +x}\;dx < \int_0^x1 \; dx \\ {x \over 1+x} < \ln(1+x) < x \]

    • 2 years ago
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    \[ \ln (x) \text{ in } [1, 1+t], t>0 \] perhaps this would mean show that \( \ln(1)<{t \over 1 +t} < \ln(1+t)\)

    • 2 years ago
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    MVT http://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/Mvt2.svg/300px-Mvt2.svg.png

    • 2 years ago
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    |dw:1350722628100:dw|

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    |dw:1350722670829:dw|

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    |dw:1350722732492:dw|

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    |dw:1350722797470:dw|

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    |dw:1350722973766:dw||dw:1350723003576:dw|

    • 2 years ago
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    |dw:1350723085914:dw|

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  20. Callisto Group Title
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    f(b) - f(a) / (b-a) ?

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    yep!!

    • 2 years ago
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    |dw:1350724369177:dw|

    • 2 years ago
  23. Callisto Group Title
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    OMG! My super smart brother!!

    • 2 years ago
  24. experimentX Group Title
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    lol ... not that smart. Just copied half from your prev answer.

    • 2 years ago
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