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Proove that e^x is always positive

  1. Feb 16, 2013 #1
    May sound trivial but I don't find it trivial from the things I am given as a definition of the exponential function:

    The exponential function satisfies f(a+b)=f(a)f(b), is the inverse to the ln and is restricted by the condition:
    exp(x)≥1+x

    I can't see how I can from this only proove that it is always positive. Can anyone help?
     
  2. jcsd
  3. Feb 16, 2013 #2

    Dick

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    If you are given exp(x)≥1+x then you already know exp(x)>0 if x≥0, right? So try and think of a way to show it if x<0.
     
  4. Feb 16, 2013 #3
    ugh yeah that is where the hard work comes in for my part.

    I have tried: exp(-x) ≥ 1 - x

    So exp(x) ≤ 1/(1-x)

    but that, as you can see, didn't get me anywhere.

    But maybe this will do: for x>0

    exp(x)exp(-x) = 1 which means exp(x) must be positive for x<0?
     
  5. Feb 16, 2013 #4

    Ray Vickson

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    Well, what do YOU think?
     
  6. Feb 16, 2013 #5
    Yes!
     
  7. Feb 16, 2013 #6

    Fredrik

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    I just want to say that there's a much easier way than what's been discussed so far.
     
  8. Feb 16, 2013 #7
    and that is?
     
  9. Feb 16, 2013 #8

    Dick

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    There's probably a lot of ways to do it. But since you have those givens, I can't think of anything much easier. BTW you weren't really given that exp(0)=1. You just know exp(0)>0, which is enough. You can, of course, prove exp(0)=1 from what you are given.
     
  10. Feb 16, 2013 #9

    Fredrik

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    We're only supposed to give hints, not complete solutions, but since you have found one solution I guess it's OK. What I had in mind is to use that 1=1/2+1/2.
    $$e^x=e^{\frac 1 2 x+\frac 1 2 x}=\left(e^{\frac 1 2 x}\right)^2\geq 0.$$ Of course, you still have to prove that ##e^x\neq 0## for all x. But this is easy. Suppose that it's not true. Let y be a real number such that ##e^y=0##. Let x be an arbitrary real number.
    $$0=e^{x-y}0=e^{x-y}e^y=e^{x-y+y}=e^x.$$ Since x is arbitrary, this contradicts that ##e^0\geq 1##.

    When I said that this is "much" easier, I didn't realize that we still had to prove that ##e^x\neq 0## after the first step.
     
  11. Feb 16, 2013 #10

    Dick

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    Sure, f(0)=0 also satisfies f(a+b)=f(a)f(b). You need some other input. f(0) not equal to 0 would have been adequate. They gave you more than you really needed.
     
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