Solve the function equation $g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0$

Question

let $g(x)\in \Bbb R$ and for any $x\in \Bbb R$ such that $$g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0, g(0)=0$$

find $g(x)$

my idea let $x\longrightarrow x+1$, then we have $$g^2(x+1)-g(x+2)-\dfrac{(x+1)^2+2(x+1)-6}{4}=0$$ $$g^2(x+2)-g(x+3)-\dfrac{(x+2)^2+2(x+2)-6}{4}=0$$ $$\cdots\cdots $$ $$g^2(x+n)-g(x+n+1)-\dfrac{(x+n)^2+2(x+n)-6}{4}=0$$

following I can't work,Thank you

One of the answers below seems to understand $,g^2(x)=(g(x))^2;$ Is this correct or you meant $,g^2(x)=g(g(x))=g\circ g(x);$ ? — DonAntonio, Aug 03 '13 at 12:16

score 5 · Answer 1 · answered Aug 03 '13 at 12:19

5

Take $\,x=-1\;$:

$$g(-1)^2-g(0)-\frac{(-1)^2+2(-1)-6}4=0\implies g(-1)^2=-\frac74$$

which of course is impossible unless there are further conditions or we're working with complex numbers and not with the reals, as written, say.

answered Aug 03 '13 at 12:19

DonAntonio

211,718
17
136
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score 0 · Answer 2 · answered Aug 03 '13 at 12:12

0

HINT:

Let $g(x)=\sum_{n\ge i\ge 0} a_ix^i$

So, $\displaystyle \frac{x^2+2x-6}4= g^2(x)-g(x+1)=(\sum_{n\ge i\ge 0} a_ix^i)^2-\sum_{n\ge i\ge 0} a_i(x+1)^i=a^2_nx^{2n}+\cdots$

So comparing the coefficients of the highest power of $x$, $n=1$ and $a_1=\frac12$

$\implies g(x)=\frac x2+a_0$

As $g(0)=0$ and $g(0)=a_0\implies \displaystyle a_0=0\implies g(x)=\frac x2$

So, $\displaystyle g^2(x)-g(x+1)=\left(\frac x2\right)^2-\frac{x+1}2=\frac{x^2-2x-2}4$

can this be equal to $\displaystyle\frac{x^2+2x-6}4$

answered Aug 03 '13 at 12:12

lab bhattacharjee

274,582

It's good, but we also must prove that the solution of equation must be only polynoms and nothing else. – cool Aug 03 '13 at 12:16
@Paul, I was just trying to identify a counter-example of : if the difference is polynomial, $g(x)$ is something else – lab bhattacharjee Aug 03 '13 at 12:19
Ah, yes. I have slightly misunderstood. – cool Aug 03 '13 at 12:21

Solve the function equation $g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0$

2 Answers2