The following expression is given, and we are asked to find $f(2)$. \begin{equation} 2f(x)-3f\left(\frac{1}{x}\right) =x^2 \end{equation}

Does a unique and well defined answer exist? Why? and what is it?

We have that $f(1)=f(-1)=-1$.

  • $\begingroup$ Note that we have no extra information, such as whether it is continuous or analytic . $\endgroup$ – Student tea Feb 20 '15 at 15:42

Make $x = 2$ and get $2f(2) - 3f(0.5) = 4$. Make $x = 0.5$ and get $2f(0.5) - 3f(2) = 0.25$. Solve the system: $$\begin{cases} 2f(2) - 3f(0.5) = 4 \\ -3f(2) + 2f(0.5) = 0.25\end{cases}.$$

  • 4
    $\begingroup$ Nice. Same approach helps express $f$ explicitly. $\endgroup$ – Macavity Feb 20 '15 at 15:51

Hint: Note that $$ \begin{bmatrix} 2&-3\\ -3&2 \end{bmatrix} \begin{bmatrix} f(x)\\ f(1/x) \end{bmatrix} = \begin{bmatrix} x^2\\ 1/x^2 \end{bmatrix} $$ and $$ \det\begin{bmatrix} 2&-3\\ -3&2 \end{bmatrix}=-5\ne0 $$

Completed Answer

Now that over a year has passed, I think it is appropriate to finish the answer. $$ \begin{align} \begin{bmatrix} f(x)\\ f(1/x) \end{bmatrix} &= \begin{bmatrix} 2&-3\\ -3&2 \end{bmatrix}^{-1} \begin{bmatrix} x^2\\ 1/x^2 \end{bmatrix}\\ &= -\frac15\begin{bmatrix} 2&3\\ 3&2 \end{bmatrix} \begin{bmatrix} x^2\\ 1/x^2 \end{bmatrix}\\ &=-\frac15\begin{bmatrix} 2x^2+3/x^2\\ 3x^2+2/x^2 \end{bmatrix}\\ \end{align} $$ Therefore, $$ f(x)=-\frac{2x^4+3}{5x^2} $$

  • $\begingroup$ To the downvoter: assuming that this is not a downvote out of spite and since my hint leads directly to a formula for $f$, I would appreciate knowing what is wrong with this answer. $\endgroup$ – robjohn Jul 10 '17 at 22:14
  • $\begingroup$ +1 to compensate for the work of downvoter. BTW you waited for an year to expand your hint into an answer, quite amazing! $\endgroup$ – Paramanand Singh Jul 11 '17 at 2:05

For a general equation of the type $af(x)+bf(1/x)=g(x)$, for some known $g$ with $x\ne 0$, the way to solve is to consider the two equations $$\begin{pmatrix} a & b\\ b & a\\ \end{pmatrix}\begin{pmatrix} f(x)\\ f(1/x) \end{pmatrix}=\begin{pmatrix} g(x)\\ g(1/x) \end{pmatrix}$$ This will give you a solution for $f(x)$ as long as $a\ne b$. If $a=b$, then the equation holds only for $\{x:g(x)=g(1/x)\}$


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