What is the Relationship Between x and n in x^n > (x+1)^n-1?

[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, a^x is only defined (as a real-valued function of real numbers) for a> 0 so the domain of x^x 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:
$$\frac{(x+1)^{n-1}}{x^{n}}=\frac{(1+\frac{1}{x})^{n}}{(x+1)}}=\frac{((1+\frac{1}{x})^{x})^{\frac{n}{x}}}{(x+1)}$$
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 $$x\to\infty$$