How can I solve this equation of functions?

Question

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?

Answer 1

$$f(x)=x^2\;,\;\;g(x)=xe^x{}{}{}{}{}{}{}{}{}{}{}$$

Answer 2

If you play around with exponents of monomials, then you're looking for $f(x) = x^n$ and $g(x) = x^m$. By your conditions, you require $n \geq 2, m \geq 1$.

Looking at the ratio $$ \frac{f(x)g'(x)}{(f(x)^2} = \frac{mx^{n+m-1}}{x^{2m}} = m x^{n-m-1},$$ you want a solution to $n-m-1 = 0$, or $n = m+1$.

For instance, $f(x) = x^3$ and $g(x) = x^2$ works. So does $f(x) = x^4$ and $g(x) = x^3$, and more generally $f(x) = x^m, g(x) = x^{m+1}$ for any $m \geq 2$.

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(When I first answered, I misread the constraint that they both cannot be monomials)

To account for them not both being monomials, it is perhaps easiest to base a response off the work above and incorporate knowledge from series expansion. For instance, for $x$ small, $e^x \approx 1 + x + x^2/2 + O(x^3)$. Correspondingly, $e^x - 1 \approx x + x^2/2 + O(x^3)$.

Then we can adjust $f(x)$ in the start of this answer to a similar function $\widetilde{f}(x)$ (or $g$ into $\widetilde{g}(x)$, but I happen to choose the former) by choosing $\widetilde{f}(x) = e^{mx} - 1$ in place of $f(x) = x^m$. (This works because for small $x$, $e^{mx} - 1 \approx x^m$).

So an infinite family of choices are $\widetilde{f}(x) = e^{mx}-1$ and $g(x) = x^{m+1}$.

Answer 3

A solution that uses no monomials is $$f(x)=e^x-x-1\qquad g(x)=e^x-1$$

The limit in question is $\frac{1}{2}$.

If you're really a stickler, you might want to try to avoid $e^x$ [given its expressability in terms of $\sin(x)$ and $\cos(x)$], and here you can modify mixedmath's answer with

$$f(x)=x^3+x^2 \qquad g(x)=x^2+x$$

Here the limit is 1.

Answer 4

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 $$ f(x) = \sum_{n=2}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \ln(1+x). $$