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Analysis: prove that ln(x) is a smooth function (i.e. infinitely differentiable)

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that f(x) is a smooth function (i.e. infinitely differentiable)

    2. Relevant equations

    ln(x) = [itex]\int^{x}_{1}[/itex] 1/t dt

    f(x) = ln(x)

    3. The attempt at a solution

    I was thinking about using taylor series to prove ln(x) is smooth but i'm strictly told to NOT assume f(x) = ln(x) beforehand.
     
    Last edited: Dec 1, 2011
  2. jcsd
  3. Dec 1, 2011 #2

    Office_Shredder

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    I'm confused. When you say you can't assume f(x)=ln(x), then how is f(x) defined?
     
  4. Dec 1, 2011 #3
    i guess i meant - we're supposed to derive that f(x) is smooth from f(x) = [itex]\int^{x}_{1}[/itex] 1/t dt, rather than f(x)=ln(x)
     
  5. Dec 2, 2011 #4

    Office_Shredder

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    Ok. Can you calculate f'(x) from that definition?
     
  6. Dec 2, 2011 #5
    f'(x) is just 1/t, so i just go about proving that 1/t is infinitely differentiable?
     
  7. Dec 2, 2011 #6

    Office_Shredder

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    That's correct, except it's 1/x not 1/t
     
  8. Dec 2, 2011 #7
    ah right. i know it's intuitive that 1/x is infinitely differentiable but what do we use to rigorously prove it? taylor series?
     
  9. Dec 2, 2011 #8

    Office_Shredder

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    Well, can you calculate the derivative of 1/x? How about the second derivative of 1/x? The third derivative?

    Once you see a pattern, think about a proof by induction
     
  10. Dec 2, 2011 #9
    thank you.
     
  11. Dec 2, 2011 #10

    Deveno

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    one caveat, which i feel i ought to mention:

    the domain of definition *matters*.

    you can define f on (0,∞), but not on any interval containing 0 (because f is undefined).

    a similar restriction holds for the derivatives.

    so it's not enough just to say f is smooth....you have to say WHERE f is smooth.
     
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