# Thread: A problem in analysis

1. ## A problem in analysis

Suppose e^(e^x)=an x^n, prove that an>=e(r ln n)^(-n) when n>1, r is a constant larger than e. Thanks

2. Originally Posted by cyclic

Suppose e^(e^x)=an x^n, prove that an>=e(r ln n)^(-n) when n>1, r is a constant larger than e. Thanks
the reason that you haven't got any response is that your question is not very clear. i don't know about other members in here but if a poster didn't bother to explain his/her question clearly, i.e.

mathematically understandable, then i would just ignore the question. your question may be understood in different ways. for example, this one might be what you meant:

suppose $e^{e^x}=a_nx^n,$ for some $x \in \mathbb{R}, \ a_n \in \mathbb{R}, \ 1 < n \in \mathbb{N}.$ then there exists a real constant $r > e,$ independent from $n$ and $x,$ such that $a_n \geq \frac{e}{(r \ln n)^n}.$

3. ## Thank you very much.

Your understand is correct. I am very sorry that I don't know how to post mathematical formula.

4. Originally Posted by cyclic
Your understand is correct. I am very sorry that I don't know how to post mathematical formula.
i think we also need to assume that $x > 0$ because otherwise it would be possible to choose $a_n < 0,$ which would disprove the problem. for example $e^{e^{x}}= -x$ has a root in the interval $(-2,-1).$

5. ## You are wrong

$e^{e^x}=a_nx^n$ is a series

6. Originally Posted by cyclic
$e^{e^x}=a_nx^n$ is a series
wow! that totally changes the problem!! so you mean this: $e^{e^x}=\sum_{n=0}^{\infty}a_nx^n.$

7. ## Yes, that is correct.

I think that the problem should be very difficult, becauce in a Chinese math forum there is no one can solve it

8. ## I post the problem again

Suppose $e^{e^x}=\sum_{n=0}^{\infty}a_nx^n$, prove that
$a_n \geq \frac{e}{(r \ln n)^n}$ when n>1, r is a constant larger than e.
Note that: NOt there exists a real constant r > e, BUT r is a constant larger than e