How can I solve this equation of functions? I need to find two functions continuous on $(-a, a)$, such that:


*

*$f(0) = 0$

*$f'(0) = 0$

*$g(0) = 0$

*$\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} \not= 0$

*The functions are not both writable as $x^n$, where n is a real number (one can be, but not both).

*The $\sin$ and $\cos$ functions are not used


How can I do this?

Let
$$ f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \sum_{n=0}^{\infty} b_{n} \, x^{n} $$
then by the given conditions, $f(0) = 0$, $f'(0) = 0$, $g(0) = 0$, it is found that $b_{0} = 0$, $a_{0} = a_{1} = 0$ and lead to
$$ f(x) = \sum_{n=1}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \sum_{n=2}^{\infty} b_{n} \, x^{n}. $$
There is a lot of freedom to choose the coefficients $a_{n}$ and $b_{n}$.
What remains is: 


*

*$\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} \not= 0$

*The functions are not both writable as $x^n$, where n is a real number (one can be, but not both).


In this view let $b_{n} = \frac{(-1)^{n+1}}{n}$ such that $g(x) = \ln(1+x)$. Now,
\begin{align}
\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} &= \lim_{x \to 0} \frac{(a_{2} \, x^{2} + \cdots + a_{n} \, x^{n} + \cdots)}{(1+x) \, \ln^{2}(1+x) }  \, \to \frac{0}{0} \\
&= \lim_{x \to 0} \frac{(2 \, a_{2} \, x + \cdots + n \, a_{n} \, x^{n-1} + \cdots)}{2  \, \frac{\ln(1+x)}{1+x} }  \, \to \frac{0}{0} \\
&= \lim_{x \to 0} \frac{(2 \, a_{2} + \cdots + n(n-1) \, a_{n} \, x^{n-2} + \cdots)}{2 \, \frac{1}{1+x} }  \, \to a_{2} \neq 0 \\
\end{align}
This allows for 
$$ f(x) = \sum_{n=2}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \ln(1+x). $$
