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Two very simple questions.

  1. Nov 20, 2004 #1
    1. For what "x"s are defined the function x^x (x is a real number).
    2. Demonstrate that x^n > (x+1)^n-1


    By the way, which tool do you use for including equations in the posts?
     
  2. jcsd
  3. Nov 20, 2004 #2

    T@P

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    First of all, to use equations you use LaTex, there are posts on it in most forums. Also, your second equation should have a "greater OR equal" sign, since for n = 1 and x = 1 you have 1 = 1
     
    Last edited: Nov 20, 2004
  4. Nov 21, 2004 #3
    Yea, i forgot to said yesterday that i want the proof for x>=2, x is natural.
     
  5. Nov 22, 2004 #4

    Alkatran

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    2^2 >= 3^2-1
    4 >= 8
    false
     
  6. Nov 22, 2004 #5

    HallsofIvy

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    In general, ax is only defined (as a real-valued function of real numbers) for a> 0 so the domain of xx is x> 0. If you are dealing with complex functions of complex numbers, I believe it is defined for all x not equal to 0.

    Alkatran: Good point, but I suspect he really meant:

    Prove that xn > (x+1)n-1 for x>= 2, n a natural number.

    Even then it is not true. x4> (x+1)3 only for x> approximately 2.7. In particular, 2.54= 39 1/16 while 3.53= 42 7/8.

    Eventually xn> (x+1)n-1 but the value of x for which that is true increases with n.
     
  7. Nov 22, 2004 #6

    arildno

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    To get a "feel" of the dependence on "n", consider the ratio:
    [tex]\frac{(x+1)^{n-1}}{x^{n}}=\frac{(1+\frac{1}{x})^{n}}{(x+1)}}=\frac{((1+\frac{1}{x})^{x})^{\frac{n}{x}}}{(x+1)}[/tex]
    For big x, we have:
    [tex](1+\frac{1}{x})^{x}\approx{e}[/tex]
    That is, for big x, our expression is approximately:
    [tex]\frac{e^{\frac{n}{x}}}{(x+1)}[/tex]
    Since the numerator goes to 1 as [tex]x\to\infty[/tex] we see that the fraction goes to zero as [tex]x\to\infty[/tex]
     
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