# A problem in analysis

• Dec 31st 2008, 04:56 AM
cyclic
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
• Jan 1st 2009, 08:39 PM
NonCommAlg
Quote:

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 $a_n \geq \frac{e}{(r \ln n)^n}.$
• Jan 2nd 2009, 03:18 AM
cyclic
Thank you very much.
Your understand is correct. I am very sorry that I don't know how to post mathematical formula.
• Jan 2nd 2009, 03:33 AM
NonCommAlg
Quote:

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 $(-2,-1).$
• Jan 2nd 2009, 05:20 AM
cyclic
You are wrong
$e^{e^x}=a_nx^n$ is a series
• Jan 2nd 2009, 05:30 AM
NonCommAlg
Quote:

Originally Posted by cyclic
$e^{e^x}=a_nx^n$ is a series

wow! that totally changes the problem!! so you mean this: $e^{e^x}=\sum_{n=0}^{\infty}a_nx^n.$
• Jan 2nd 2009, 05:39 AM
cyclic
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
• Jan 2nd 2009, 05:51 AM
cyclic
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