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Let $f$ be a function $f:\mathbb R\to\mathbb R$.
Find all functions $f$ that satisfy:
$$f(x^2+x+3)+2f(x^2-3x+5)=x^2-x+ \frac{18}{4} + \frac{111}{444} + \frac{222}{333}$$
Maybe the question is wrong; I found it on a forum on the internet without any answers.
Note: The question has no solutions, I'm sorry. On the forum someone says he made it so probably there are no functions that satisfy this.
I continue from Christian's work.
Define $h(t) = g(t^2+2t)$ (so that $g(t)$ may be solved for by inverting the quadratic and taking suitable care). Then we have $$h(t) + 2h(-t) = t^2$$
Assuming we are interesting in $h$ which are infinitely differentiable at the origin, we can evaluate the $k$th derivative at the origin to obtain $$h^{(k)}(0)(1+2(-1)^k ) = 2\delta_{k,2}$$ or $$h^{(k)}(0) = \begin{cases} 0 & k\neq2 \\ \frac2 3 & k=2 \end{cases}$$ and so if we further consider functions which are analytic at the origin, we find $$h(t) = \frac 1 3 t^2$$
Then clearly if $x=t^2+2t$ we have $$t^2+2t-x=0 \iff t = -1 \pm \sqrt{1+x}$$
Hence $$g(x) = h(-1\pm\sqrt{1+x}) = \frac 1 3(\sqrt{1+x}\mp 1)^2=\frac 1 3 (2+x\mp 2\sqrt{1+x})$$
Finally, checking which signs work where, I find that $$\boxed{f(t)=\frac 1 {36} (41+12t-12\sqrt{4t-11}) \qquad\text{makes relation hold for}\qquad x \in \left(-\frac 1 2, \frac 3 2\right)}$$
The range has two ends because $f$ is evaluated at two places, and the solution breaks down where the sign changeover arises.
Note: This $g$ has series expansion agreeing with Christian's.
A beginning:
Write $x:={1\over2}+t$. Then your functional equation goes over into $$f\bigl({15\over4}+2t+t^2\bigr)+2f\bigl({15\over4}-2t+t^2\bigr)=t^2+{31\over6}\ .$$ Introduce a new unknown function $g$ by means of $$g(\tau):=f\bigl({15\over4}+\tau\bigr)-{31\over18}\ .$$ Then $g(0)=0$, and $g$ satisfies the functional equation $$g(2t+t^2)+2g(-2t+t^2)= t^2\ .$$ Writing $g(t)=\sum_{k=0}^\infty a_k t^k$ and comparing coefficients one obtains $$\eqalign{g(t)&={t^2\over12} - {t^3\over24} + {5 t^4\over192} - {7 t^5\over384} + {7 t^6\over512} - { 11 t^7\over1024} + {143 t^8\over16384} - {715 t^9\over98304} +\cr&\qquad { 2431 t^{10}\over393216} - {4199 t^{11}\over786432} + {29393 t^{12}\over6291456} - { 52003 t^{13}\over12582912} + {185725 t^{14}\over50331648}+\ldots\cr}$$ The series so produced seems to have a radius of convergence $>1$. At any rate one would have to look at the exact recursion formula for the coefficients.
It happens that this question is pure algebra...
No solution $f$ (whatever its regularity) can be defined on the five points $\frac{11}4$, $\frac{11}4\pm4$, $\frac{11}4\pm16$.
Consider the identity to be solved when the RHS is $x^2-x+c$, for some $c$ whose value will be irrelevant, and let $a=\tfrac34+c$. Some tedious but easy algebraic computations show that the sequence $(x_n)$ defined by $$ x_n=f\left(4n^2+\tfrac{11}4\right) $$ solves the identities $$ x_n+2x_{n-1}=4n(n-1)+a. $$ For example, $x_1=a-2x_0$, $x_2=8-a+4x_0$, $x_{-1}=\frac12a-\frac12x_0$ and $x_{-2}=4+\frac14a+\frac14x_0$.
By definition, $x_{-n}=x_n$ for every $n$, in particular, $$ a-2x_0=\tfrac12a-\tfrac12x_0,\qquad 8-a+4x_0=4+\tfrac14a+\tfrac14x_0. $$ The first identity implies that $a=3x_0$, and the second identity reads $4+\frac{15}4x_0=\frac54a$, which has no solution when the first identity holds since then, $\frac{15}4x_0=\frac54a$.
