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Identity for cosh

  1. Feb 14, 2013 #1
    Show that cosh(x) = 1 => x = 0

    I am only allowed to use the definition of cosh, the algebraic rules for the exponential function, that exp(x2)>exp(x1) for x2 > x1, and the fact that we have defined it with the requirement:

    exp(x) ≥ 1 + x

    The exp(x) term of course is not trouble.

    What I lack prooving is that:
    exp(x) > 1 - x

    Any ideas how I might approach it? I don't need answers, just hints
     
  2. jcsd
  3. Feb 14, 2013 #2

    CompuChip

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    If x > 0, then 2x > 0 so 1 + x > 1 + x - 2x.

    If x < 0, then it's not true (exp goes to 0 while 1 - x goes to infinity for [itex]x \to -\infty[/itex]).
     
  4. Feb 14, 2013 #3
    oh sorry I can see I made a fatal error.
    I meant i need to show:
    exp(-x) > 1 - x
     
  5. Feb 14, 2013 #4

    Mark44

    Staff: Mentor

    A quick sketch of the graphs of y = e-x and y = 1 - x shows that e-x ≥ 1 - x for all real x, and that we get equality at only one value of x.

    Assuming you can use techniques of calculus, let f(x) = e-x - (1 - x), and find the minimum value and where it occurs.
     
  6. Feb 14, 2013 #5
    I know calculus but this is an analysis course, where it has not yet been introduced. Thus I am to find another way of proving it.
     
  7. Feb 14, 2013 #6

    Mark44

    Staff: Mentor

    Really? Are you saying that you haven't learned differentiation, and you're in an analysis course?
     
  8. Feb 14, 2013 #7
    I'm talking a rigorous analysis course where everything is built up slowly. So far we have covered continuity and the properties of limits. I know differential calculus, but I am not allowed to use it for this handout.
     
  9. Feb 14, 2013 #8

    vela

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    If you know that ##e^x \ge 1+x## for all ##x##, you can plug anything you want in for ##x##. In particular, you can plug in ##-x##.
     
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