Infinite Fraction

[Ceby]

I need some help with this. I calculated it using computer and it converges to 1.73202714. I'm not so sure about it because when I compute I assumed that if x become LARGE, then f(x)=1. Is my assumption correct? and is there a way to find the answer without using computer?

 

[Ishuda]

I need some help with this. I calculated it using computer and it converges to 1.73202714. I'm not so sure about it because when I compute I assumed that if x become LARGE, then f(x)=1. Is my assumption correct? and is there a way to find the answer without using computer?

If f(x) is going to approach some sort of finite limit as x becomes very large, then, for x 'very large', f(x), f(2x) and f(3x) would have to be about the same and f(2x)/f(3x) would be about 1. That would imply that f(x) [= x + f(2x)/f(3x)] would be about x + 1 and would increase without bound and thus it can not approach a finite limit.

So suppose f were of polynomial growth in the sense that
f(x) ~ a xⁿ as x approached infinity.
That is suppose there is an a > 0
f(x)/xⁿ = a as x approached infinity.
Then f(2x)/(2x)ⁿ $\displaystyle \to$ a and f(3x)/(3x)ⁿ $\displaystyle \to$ a as x $\displaystyle \to \infty$
or
f(2x)/f(3x) = [f(2x)/(2x)ⁿ] [ (3x)ⁿ/f(3x) ] (2/3)ⁿ $\displaystyle \to$ (a) (1/a) (2/3)ⁿ = (2/3)ⁿ as x $\displaystyle \to \infty$
and
f(x) $\displaystyle \to$ x + (2/3)ⁿ
and a and n are equal to 1.

Note that that is proved only if f(x) is of polynomial growth. Do you have any other information on f?

EDIT: Also notice that the above does not mean that f(x) is a polynomial. For example consider the following with $\displaystyle \alpha\, and\, \beta$ real and $\displaystyle \beta \gt$ 0:
f(x) = x + (2/3) [ 1 - $\displaystyle \alpha\, e^{-\beta\, x}$]