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?
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Infinite Fraction
- Thread starter Ceby
- Start date
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 - \(\displaystyle \alpha\, e^{-\beta\, x}\)]
Last edited: