Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I can easily show that with the assumption $f$ is a polynomial $f(x)=x^2$. But without that assumption how can I prove that $f(x)=x^2$???. I have tried many change of variables $x=u+k$ but to no result. I am lost here

share|improve this question

3 Answers 3

up vote 6 down vote accepted

Let $g(x)$ be so that $3g(x)-2g(x-3)=0$. For example, take any periodic $h(x)$ on $[0,1]$ and let $g(x)=h(x/3)\left(\frac23\right)^{x/3}$. Then $g(x)+x^2$ also satisfies your equation. So there are many solutions to your equation.

share|improve this answer
    
+1 nice...I was way off track! –  amWhy Nov 11 '12 at 14:33

The following note - for a general functional equation which is related

Assume that $F=\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$ and $V$ and $B$ are vector spaces over $F$ and $X$ is a Banach spaces over $F$.

A function $a:V\rightarrow B$ is said to be additive provided $a(x+y)=a(x)+a(y)$ for all $x, y\in V$; in this case it is easily seen that $a(rx) =ra(x)$ for all $x\in V$ and all $r\in \mathbb{Q}$ [1].

If $k\in \mathbb{N}$ and $a:V^{k}\rightarrow B$, then we say that a is $k$-additive provided it is additive in each variable; we say that a is symmetric provided $a(x_{1},x_{2},...,x_{k})=a(y_{1},y_{2},...,y_{k})$ whenever $x_{1},x_{2},...,x_{k}\in V$ and $(y_{1},y_{2},...,y_{k})$ is a permutation of $(x_{1},x_{2},...,x_{k})$.

If $k\in \mathbb{N}$ and $a:V^{k}\rightarrow B$ is symmetric and $k$-additive, let $a^{\ast}(x)=a(x,x,...,x)$ for $x\in V$ is and note that $a^{\ast}(rx)=r^{k}a(x)$ whenever $x\in V$ and $r\in \mathbb{Q}$. Such a function $a^{\ast}$ will be called a monomial function of degree $k$ (assuming $a^{\ast}\neq 0$).

A function $p:V\rightarrow B$ is called a generalized polynomial} (GP) function of degree $m\in \mathbb{N}$ provided

there exist $a_{0}\in B$ and symmetric $k$-additive functions $a_{k}:V^{k}\rightarrow B$ (for $1\leq k\leq m$) such that $$p(x)=a_{0}+\sum_{k=1}^{m}a_{k}^{\ast}(x)\ \ \ for\ all\ x\in V,$$ and $a^{\ast}_{m}\neq 0$. In this case $$p(rx)=a_{0}+\sum_{k=1}^{m}r^{k}a_{k}^{\ast}(x)\ \ \ for\ all\ x\in V\ and\ r\in \mathbb{Q}.$$

Let $B^{V}$ denote the vector space (over $F$) consisting of all maps from $V$ into $B$. For $h\in V$ define the linear difference operator $\Delta_{h}$ on $B^{V}$ by $$\Delta_{h}f(x)=f(x+h)-f(x)\ \ \ for\ all\ f\in B^{V} and\ x\in V.$$ Notice that these difference operators commute ($\Delta_{h_{1}}\Delta_{h_{2}}=\Delta_{h_{2}}\Delta_{h_{2}}$ for all $h_{1},h_{2}\in V$) and if $h\in V$ and $n\in \mathbb{N}$, then $\Delta_{h}^{n}$--the $n$-th iterate of $\Delta_{h}$--satisfies $$\Delta_{h}^{n}f(x)=\sum_{k=0}^{n}(-1)^{n-k}(_{k}^{n})f(x+kh)\ \ \ for\ f\in B^{V}\ and\ x,h\in V.$$

The following theorem were proved by Mazur and Orlicz [2] and [3], and in greater generality by Djokovi$\acute{c}$ [4].

Theorem:

If $n\in \mathbb{N}$ and $f:V\rightarrow B$, then the following are equivalent.

$\Delta_{h}^{n}f(x)=0$ for all $x, h\in V$.

$\Delta_{h_{n}}...\Delta_{h_{1}}f(x)=0$ for all $x,h_{1},...,h_{n}\in V.$

$f$ is a GP function of degree at most $n-1$.

for more information see http://arxiv.org/abs/1210.4975v1

share|improve this answer
    
With the above theorem can we proved that $f$ is a monomial function of degree 2. –  Alisad Nov 11 '12 at 15:15

Let $f(x)$ be polynomial of $x$.

Let $n\ge 3,$ be the smallest power of $x$ whose coefficient$(a)$ may be $\ne 0$.

The coefficient of $x^n$ in $3f(x−6)−2f(x−9)$ is $3a-2a=a$

But the coefficient of $x^n$ where $n\ge 3$ in $x^2−54$ is $0\implies a=0$

So, $f(x)$ can not contain any higher powers $(\ge 3)$ of $x.$

So, $f(x)=bx^2+cx+d$

$3f(x−6)−2f(x−9)=x^2−54$

$b(x-h)^2+c(x-h)+d=x^2(b)+x(c-2bh)+(d-ch+bh^2)$

The coefficient of $x^2$ in $3f(x−6)−2f(x−9)$ is $3b-2b=b$

But the coefficient of $x^2$ in $x^2−54$ is $1\implies b=1$

The coefficient of $x$ in $f(x−h)$ is $c-2bh=c-2h$

So, the coefficient of $x$ in $3f(x−6)−2f(x−9)$ is $3\{c-2(-6)\}-2\{c-2(-9)\}=c$

But the coefficient of $x$ in $x^2−54$ is $0\implies c=0$

The constant term in $f(x−h)$ is $d-ch+bh^2=d+h^2$

The constant term in $3f(x−6)−2f(x−9$ is $3\{d+(-6)^2\}-2\{d+(-9)^2\}=d-54$

The constant term in $x^2−54$ is $-54\implies d-54=-54\implies d=0$

So, $f(x)=x^2+0x+0=x^2$

share|improve this answer
    
+@lab nice work! –  amWhy Nov 11 '12 at 14:50
    
@amWhy, thank, please have a look into the edited answer. –  lab bhattacharjee Nov 11 '12 at 14:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.