Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f^{k+1}(x)=f(f^k(x))$ and $f^0(x)=f(x)>x$ such that $\displaystyle{\frac{df}{dx}(x)>0}$ for all $x>0$. Does it imply that $$g(x)=\sum_{k=1}^\infty \frac{1}{f^k(x)}<\infty$$ for all $x>0$?

share|improve this question
2  
Usually one takes $f^0(x)=x$ and $f^1(x)=f(x)$. –  Michael Hardy Nov 16 '11 at 16:43

3 Answers 3

up vote 6 down vote accepted

No. Here’s an even easier example. Take $f(x)=x+1$; then $f^k(x)=x+k+1>x$, and $f\;'(x)=1$ for all $x$. But $$\sum_{k\ge 1}\frac1{f^k(x)} = \sum_{k\ge 2}\frac1{x+k},$$ which clearly diverges for $x>0$.

share|improve this answer

No. Consider $f(x)=x+c/(x+1)$ with $0<c\leqslant1$. Then $f'(x)\gt1-c\geqslant0$ for every $x\gt0$ and $f(x)^2=x^2+2c+o(1)$ when $x\to+\infty$ hence, for every fixed $x\gt0$, $[f^k(x)]^2=2ck+o(k)$ when $k\to\infty$. In particular, $1/f^k(x)$ is of order $1/\sqrt{k}$ hence $g(x)$ is infinite for every $x\gt0$.

share|improve this answer

No. If there is no typo in the question, take $f(x) = x$ (then $f^k(x) = x$ independently of $k$).

share|improve this answer
1  
Sorry for typo. its f(x)>x –  100101011011010011010110101110 Nov 16 '11 at 16:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.