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Le^x-(1+x/1!+x^2/2!)l <= e/6 for all x E [0,1].

  1. May 5, 2009 #1
    I have a problem.
    How do I show that le^x-(1+x/1!+x^2/2!)l <= e/6 for all x E [0,1].
  2. jcsd
  3. May 5, 2009 #2
    Re: e^x

    Show that the equation is true at the boundries. Show that the term on the left is monotonic in x (if, indeed, it is).
  4. May 6, 2009 #3
    Re: e^x

    What a strange coincidence. I saw this thread only after posting my example in this thread.

    If you don't want to see the full solution, here's a hint: Taylor's theorem.
  5. May 8, 2009 #4
    Re: e^x

    how to show that a function is monotonic increasing? show that f(x+1) >= f(x) for all x?
  6. May 8, 2009 #5


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    Gold Member

    Re: e^x

    No, that won't work. A function is monotonic increasing if f(x) ≥ f(y) whenever x ≥ y.
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