# Two very simple questions.

1. Nov 20, 2004

### 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?

2. Nov 20, 2004

### 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

Last edited: Nov 20, 2004
3. Nov 21, 2004

### mprm86

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

4. Nov 22, 2004

### Alkatran

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

5. Nov 22, 2004

### 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.

6. Nov 22, 2004

### arildno

To get a "feel" of the dependence on "n", consider the ratio:
$$\frac{(x+1)^{n-1}}{x^{n}}=\frac{(1+\frac{1}{x})^{n}}{(x+1)}}=\frac{((1+\frac{1}{x})^{x})^{\frac{n}{x}}}{(x+1)}$$
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 $$x\to\infty$$