6
$\begingroup$

Someone gave me a random maths problem to solve:

Given that $h \left ( \dfrac{x}{x^2+h(x)} \right )=1$, what is $h(x)$

The restrictions given were:

  • $h(x) \neq constant$
  • $\exists \frac{dh}{dx}$
  • $\exists h^{-1}(x)$
  • $\exists \frac{dh^{-1}}{dx}$

However, I am not sure how to go about solving this. I started off with:

$$\begin{align*} \frac{x}{x^2+h(x)} &= h^{-1}(1)\\\\ \frac{x^2+h(x)}{x} &= \frac{1}{h^{-1}(1)}\\\\ x^2+h(x) &= \frac{x}{h^{-1}(1)}\\\\ h(x) &= \frac{x}{h^{-1}(1)} - x^2 \end{align*}$$

which doesn't feel like it gets me any closer to something that makes sense. I don't even know what keywords to do some Google searching with in order to educate myself on the kind of manipulations that can be done to solve this kind of question (searching for "functions that call themselves" or "recursive function" generate tons of what look like entirely unrelated results).

What do I need to know in order to be able to tackle this kind of problem?

$\endgroup$
17
  • 2
    $\begingroup$ Certainly $h(x) = 1$ is such a function. $\endgroup$ Commented Jul 25, 2015 at 20:12
  • 1
    $\begingroup$ I am assuming your function goes from and to $\mathbb{R}$. Anyway, you can't take the inverse $h^{-1}(1)$ because $h$ may not be one-to-one. For instance, the function pointed out in the comment above mine is not one-to-one. $\endgroup$
    – balddraz
    Commented Jul 25, 2015 at 20:17
  • 1
    $\begingroup$ what else is given? $\endgroup$ Commented Jul 25, 2015 at 20:19
  • $\begingroup$ Put $x = h^{-1}(1)$ to find that $ h^{-1}(1) = \pm i$ $\endgroup$ Commented Jul 25, 2015 at 20:34
  • $\begingroup$ @CountIblis It should be noted that your calculation is valid GIVEN you assume only a single $x$ is mapped to $1$ by $h$. The function pointed out in the beginning ($h(x) = 1$) already contradicts your suggestion (since $h^{-1}(1) = \text{dom} h$ which is not necessarily just $\{+i,-i\}$). $\endgroup$
    – balddraz
    Commented Jul 25, 2015 at 20:41

2 Answers 2

4
$\begingroup$

Assuming $h : \mathbb{R} \to \mathbb{R}$, there is trivially no such $h$. Since $h^{-1}$ exists, you correctly deduced that $$ \frac{x}{x^2 + h(x)} = h^{-1}(1) \;\; \forall \;x. $$ Plugging in $x = 0$, $h^{-1}(1) = 0$. But plugging in $x \ne 0$, a nonzero number over a real number then equals zero, contradiction.

EDIT: In fact, the exact same argument shows there is no $h$ defined for $x = 0$ and at least one other complex number.

$\endgroup$
12
  • $\begingroup$ wouldn't that only hold if $h(0)=0$, which has not been established? There might be a constant in the expression of $h(x)$ for instance, such that $h(0) \neq 0$? $\endgroup$ Commented Jul 25, 2015 at 21:06
  • $\begingroup$ @Mike'Pomax'Kamermans no, it's just because a nonzero number over a real number is still nonzero. $\endgroup$ Commented Jul 25, 2015 at 21:09
  • $\begingroup$ There is a serious step you are missing: when you plug in $x = 0$, you have to make sure that $h(0) \neq 0$ or else $\frac{x}{x^2 + h(x)} = \frac{0}{0}$ which is an undefined quantity and could be anything (it may be $0$, it may not be $0$). $\endgroup$
    – balddraz
    Commented Jul 25, 2015 at 21:12
  • $\begingroup$ @ZeroXLR No. The problem statement uses $\frac{x}{x^2 + h(x)}$, implying it is defined. In particular, we must have $x^2 + h(x) \ne 0$ for all $x$. If the functional equation is not valid for all $x$, then the problem statement should be clarified. $\endgroup$ Commented Jul 25, 2015 at 21:14
  • 2
    $\begingroup$ @Mike'Pomax'Kamermans The same solution works if $h: X \to \mathbb{C}$ if $X$ is a set containing $0$ and at least one other number. In particular there is no such $h: \mathbb{C} \to \mathbb{C}$. $\endgroup$ Commented Jul 25, 2015 at 21:31
0
$\begingroup$

Given $$ h\left( \frac{ x }{ x^2 + h(x) } \right) = 1. $$

Then $$ \frac{ x }{ x^2 + h(x) } = h^{-1}(1). $$

Whence

$$ h(x) = \frac{x}{h^{-1}(1)} - x^2. $$

$\endgroup$
3
  • 2
    $\begingroup$ Plug in $x = h^{-1}(1)$ to that equation and you get $1 = 1 - h^{-1}(1)^2$, i.e. $h^{-1}(1) = 0$. So $h(x) = \frac{x}{0} - x^2$. So there is no such $h$. $\endgroup$ Commented Jul 25, 2015 at 21:21
  • 2
    $\begingroup$ That's not a solution yet (you'll find that formula already mentioned in the question). $\endgroup$ Commented Jul 25, 2015 at 21:22
  • $\begingroup$ @6005 - indeed... $\endgroup$ Commented Jul 25, 2015 at 22:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .