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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?

Mike 'Pomax' Kamermans
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    Certainly $h(x) = 1$ is such a function. – Travis Willse Jul 25 '15 at 20:12
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    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. – balddraz Jul 25 '15 at 20:17
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    what else is given? – Dr. Sonnhard Graubner Jul 25 '15 at 20:19
  • Put $x = h^{-1}(1)$ to find that $ h^{-1}(1) = \pm i$ – Count Iblis Jul 25 '15 at 20:34
  • @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 ${+i,-i}$). – balddraz Jul 25 '15 at 20:41
  • @ZeroXLR CountIblis calculation gives me $h^{-1}(1)=0$ (assuming only a single $x$ is mapped onto $1$). – Mirko Jul 25 '15 at 20:44
  • Ah yes! That is true. So his suggestion is not valid even under those conditions. Well, my main point still stands. – balddraz Jul 25 '15 at 20:47
  • apologies, I forgot to add that h(x) could not be constant, so h(x)=1 was forbidden (presumably because it makes for a more interesting problem to give to someone) – Mike 'Pomax' Kamermans Jul 25 '15 at 20:55
  • the restrictions that $h$ is not a constant does not provide a unique solution, for example $h(x)=2$ if $x>1$, and $h(x)=1$ if $x\le1$ works, and in this definition you could replace $2$ with any number $>1$ and have another non-constant solution. Is $h$ supposed to be continuous? – Mirko Jul 25 '15 at 20:55
  • The answer is supposedly universal, so no partial solutions for different intervals of x. I've added all the stipulations I was given to the question. – Mike 'Pomax' Kamermans Jul 25 '15 at 20:58
  • could you clarify the meaning of the word "universal" in this context please? I have a piecewise defined function, that does not make it a partial solution, it is an honesty-good solution for the problem as stated. Hah, well, not after you added the condition that $h$ is differentiable. – Mirko Jul 25 '15 at 21:00
  • Just clarifying: by $\exists h^{-1} (x)$, you mean the function is bijective right? – balddraz Jul 25 '15 at 21:01
  • What is the domain on which $h$ is defined? – Caleb Stanford Jul 25 '15 at 21:01
  • you could start with $h(0)\not=0$ (for else $1=h(\dfrac{0}{0^2+h(0)})=h(\dfrac{0}{0})$ which is undefined) hence $h(0)=h(\dfrac{0}{0^2+h(0)})=1$. – Mirko Jul 25 '15 at 21:03
  • There is supposed to be a solution that works "for any and all x". As for the domain, this was not specified, so I assume $\mathbb{R}$ but won't rule out $\mathbb{C}$. As for the inverse function, I read the stipulations as "h(x) is continuously differentiable" and "there is an inverse, and it too is continuously differentiable", and assume it means the inverse is bijective – Mike 'Pomax' Kamermans Jul 25 '15 at 21:03
  • @Mirko this feels more like playing with inputs rather than doing the math I'm sure my friend is hoping I am able to perform. This gives us "numbers", which his stipulation $h(x) \neq constant$ rules out as acceptable. Given $h(0)=1$, how would I go about finding the function expression of $h(x)$? – Mike 'Pomax' Kamermans Jul 25 '15 at 21:12
  • I would perhaps have kept thinking about it, but @6005 already posted an answer that no such $h$ exist (after the additional conditions that you posted that $h$ is one-to-one etc) and I see no errors in his/her answer. – Mirko Jul 25 '15 at 21:22

2 Answers2

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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.

Caleb Stanford
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  • 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 $h(0) \neq 0$? – Mike 'Pomax' Kamermans Jul 25 '15 at 21:06
  • @Mike'Pomax'Kamermans no, it's just because a nonzero number over a real number is still nonzero. – Caleb Stanford Jul 25 '15 at 21:09
  • 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 $0$). – balddraz Jul 25 '15 at 21:12
  • @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. – Caleb Stanford Jul 25 '15 at 21:14
  • As far as the stipulations suggest, the function is differentiable for all $x$, including $x=0$ – Mike 'Pomax' Kamermans Jul 25 '15 at 21:15
  • @ZeroXLR I second 6005's opinion, as I had already deduced that $h(0)=1$ in one of my comments to the question. – Mirko Jul 25 '15 at 21:15
  • @Mirko Ah yes. I checked back and re-saw that comment. Indeed, in that case, I rest my case. – balddraz Jul 25 '15 at 21:18
  • This does assume $h : \mathbb{R} \to \mathbb{R}$, though. I'm fairly certain my far-more-competent-at-maths-than-I friend has a solution in mind, so $h : \mathbb{C} \to \mathbb{C}$ might have to be assumed if there is no solution in $h : \mathbb{R} \to \mathbb{R}$ – Mike 'Pomax' Kamermans Jul 25 '15 at 21:24
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    @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 $h: \mathbb{C} \to \mathbb{C}$. – Caleb Stanford Jul 25 '15 at 21:31
  • That is assuming that the functional equation is true for all $x$ where $h(x)$ is defined. So, you need to get your far-more-competent-at-maths-than-I-friend to clarify the problem, ideally she/he could put it in writing. – Caleb Stanford Jul 25 '15 at 21:31
  • yeah I think at this point I'll have to challenge him to post an answer here, so we can all see what he had in mind. Thanks for the help so far! – Mike 'Pomax' Kamermans Jul 25 '15 at 22:28
  • joke's on me, after holding him to account, it turns out his intention was to make me deduce what @6005 concluded: no such function exists. – Mike 'Pomax' Kamermans Jul 26 '15 at 18:03
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Given $$ h\left( \frac{ x }{ x^2 + h(x) } \right) = 1. $$

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

Whence

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

johannesvalks
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