Two very simple questions.

[mprm86]

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?

[T@P]

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

[mprm86]

Yea, i forgot to said yesterday that i want the proof for x>=2, x is natural.

[Alkatran]

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?

2^2 >= 3^2-1
4 >= 8
false

[HallsofIvy]

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.

[arildno]

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