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Given a polynomial $f(x)$ of n degree such that

$$f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$$

Find the polynomial

I've tried considering $f(x)=\sum \limits_{i=1}^{n} a_{i}x^{i}$ and after a number of steps I finally arrived at the answer.However I am not convinced.Is there a better way to solve this functional equation.

Thanks in advanced!

Community
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Shaswata
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  • u could just observe that f(x)=0 satisfies that, but ur solution is probably more formal :/ – Plom Aug 21 '13 at 10:40
  • Hint: Look at the degrees. – Martin Brandenburg Aug 21 '13 at 10:40
  • @Plom: It is trivial that $f=0$ is a solution. The exercise is to determine all solutions. – Martin Brandenburg Aug 21 '13 at 10:41
  • How could this ever work? How do you define $f\left(\frac{1}{0}\right)$? – Git Gud Aug 21 '13 at 10:47
  • @GitGud True, but then again, this functional equation itself is undefined for $x=0$. However, if $f(x) = 0$ for all $x$, then for any $x \neq 0$, you have $0 + 0 = 0\cdot 0$, and thus the equation is statisfied. – Arthur Aug 21 '13 at 10:49
  • @Arthur I guess my point is that a polynomial can't be a solution, at least not if my understanding of what a polynomial is, (should be defined in $\Bbb R$). I hate it when people don't use quantifiers. – Git Gud Aug 21 '13 at 10:50
  • @GitGud We can look for polynomials such that the functional equation holds for all $x\in\mathbb R\setminus{0}$ ... – Hagen von Eitzen Aug 21 '13 at 10:53
  • @HagenvonEitzen Yes, of course, hence my comment about the lack of quantifiers. – Git Gud Aug 21 '13 at 10:54

5 Answers5

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Hint: $$f(x) = 1 + \frac{1}{f(\frac{1}{x})-1}$$

So $f(x) - 1 = \frac{x^n}{P(x)}$, where P(x) is a polynomial of degree $\le n$. The left-hand side is a polynomial, so $P(x) = a x^k$ for some $k$. Now $a$ and $k$ can be found from the functional equation itself. But, as Barry Cipra notes in comments, since the left-hand side of the equality that defines $P$ has degree $n$ the only possible value for $k$ is $0$.

njguliyev
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  • nice!, can we make use of derivative to find f(x)? – Vikram Aug 21 '13 at 11:05
  • @Vikram, there is no need for derivatives. I added more details. – njguliyev Aug 21 '13 at 11:11
  • Remember that $P(x)$ can have any sign, though. So as your answer stands, you only have half the solutions. – Arthur Aug 21 '13 at 11:39
  • I think you want to say that the left hand side in the (hidden box) equation is a polynomial of degree $n$ and therefore the $P(x)$ on the right must be a constant (and then, as @Arthur notes, argue the constant is $\pm1$). – Barry Cipra Aug 21 '13 at 11:45
  • @BarryCipra, actually I thought about finding the degree and the coefficient of $P$ from the equation, but your reasoning is better. – njguliyev Aug 21 '13 at 13:04
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Assume $f(x)=c$ is constant. Then we arrive at $c+c=c\cdot c$, i.e. $c=0$ or $c=2$.

Otherwise write $f(x)=c+x^kg(x)$ where $k\ge1$ and $g(0)\ne0$. Then for $x\ne0$ $$2c+x^kg(x)+x^{-k}g(x^{-1})=f(x)+f(\frac1x)=f(x)f(\frac1x)=cx^kg(x)+c^2+cx^{-k}g(x^{-1})+g(x)g(\frac1x)$$ From looking at the highest power occuring in $g$ we see $c=1$, hence this simplifies to $$ g(x)g(\frac1x)=1$$ so $g$ is a monomial, hence constant. Thus $f(x)=1+ax^k$ and we further check that $a=\pm1$. Thus the complete list of solutions is $$ f(x)=0,\qquad f(x)=2,\qquad f(x)=1\pm x^k\text{ with }k\ge1$$

VIVID
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Hagen von Eitzen
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$$f(y) = 1+y$$

$$f(x)+f\left(\frac{1}{x}\right) = 1+x+1+\frac{1}{x} = 2 + x + \frac{1}{x}$$

$$f(x) f\left(\frac{1}{x}\right) = (1+x) \left ( 1+\frac{1}{x}\right) = 1 + x + \frac{1}{x} + x \frac{1}{x} = 2 + x + \frac{1}{x}$$

In fact, $f(y) = 1+y^k$ for $k \in \mathbb{N}$ seems to work for the same reason.

Ron Gordon
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  • $f(x)=1+x$ is only one possible solution. – Shaswata Aug 21 '13 at 10:49
  • The general solution is $f(x)=1\pm x^{n}$ – Shaswata Aug 21 '13 at 10:51
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    @shaswata: Your question seems to ask for an $f$ that satisfies the functional equation. Do you require proof that there are no other $f$? Not sure what you are looking for; the family of $f$ I have shown (with the plus/minus you point out) seems to be it, but that's only intuition. – Ron Gordon Aug 21 '13 at 10:54
  • Can you prove that no other f satisfies the above condition?Was your answer only a guess.If not could you provide the solution. – Shaswata Aug 21 '13 at 10:55
  • @shaswata: You noted in the title fin $f(x)$, not the general form of the function satisfying....! – Mikasa Aug 21 '13 at 10:56
  • @shaswata: I am not prepared to do this right now; I was only prepared to answer your original question. – Ron Gordon Aug 21 '13 at 10:56
  • i guess u could say that f(x)=/= 1+-x^n for every x and go backwards – Plom Aug 21 '13 at 11:07
  • I would personally interpret, "If $f(x)$ satisfies property $P$, then find $f(x)$" to be asking for all solutions to $f(x)$. – Alraxite Aug 21 '13 at 11:07
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A slight variation of @detnvpp using the fundamental theorem of algebra. Set $g(x)=f(x)-1$ then $$g(x)g\left(\frac{1}{x}\right)=(f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=f(x)f\left(\frac{1}{x}\right)-f(x)-f\left(\frac{1}{x}\right)+1=1, \forall x\ne 0 $$

if $\alpha \ne 0$ is a zero of the polynomial $g(x)$ then $g(\alpha)g(\frac{1}{\alpha})=0$. That contradicts $g(x)g\left(\frac{1}{x}\right)=1$. Therefore $g(x)$ has no zeros and is constant or $g(x)$ has exactly one zero $\alpha=0$ for. In the first case $g(x)=\pm 1$ and therefore $f(x)=0$ or $f(x)=2$. In the second case $g(x)=x^n$ because of $\left(g(1)\right)^2=1$ and therefore $f(x)=1 \pm x^n$

miracle173
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Suppose that the degree of $f$ is greater or equal to $1$. If $g(x)=f(x)-1$, then $g$ is a polynomial, and $$g(x)g\left(\frac{1}{x}\right)=(f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=f(x)f\left(\frac{1}{x}\right)-f(x)-f\left(\frac{1}{x}\right)+1=1.$$ Then $g(0)=0$: this can be seen from the equation above after taking the limit as $x\to\infty$. So, $\frac{g(x)}{x}$ is a polynomial that satisfies the same equation, with degree one less than the degree of $g$. So, if $g$ is of degree $n$, by induction you have that $\frac{g(x)}{x^n}$ should be constant, and equal to $\pm1$. So, $f(x)=1\pm x^n$.

detnvvp
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