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The problem that I have to solve is:

If the following function is valid for every value of $x$

$$f(3x + 1) = 9x^2 + 3x$$

find the function $f(x)$ and prove that for every $x\in\mathbb R$ the following is valid: $$f(2x) - 4f(x) = 2x$$

3 Answers 3

6

Hint: $3x+1=y \Rightarrow x= \frac{y-1}{3}$. So replace $x$ by $\frac{y-1}{3}$ and magic happens.

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$\rm \dfrac{f(3x\!+\!1)}{3x\!+\!1} = 3x\:\Rightarrow \dfrac{f(z)}z = z\!-\!1\:\Rightarrow \dfrac{f(2x)\!-\!4f(x)}{2x} = \color{#0A0}{\dfrac{f(2x)}{2x}}- 2 \color{#C00}{\dfrac{f(x)}x} = \color{#0A0}{2x\!-\!1} - 2(\color{#C00}{x\!-\!1}) = 1$

Remark $\ $ The point of presenting it like this is to emphasize how exploiting the innate linear structure serves to simplify the calculations (from nonlinear to linear). In less trivial problems this can yield much greater simplifications (e.g. in operator calculus with $q$-difference operators).

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    Why would this be downvoted?2012-08-09
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    @Downvoter If something is not clear then please feel welcome to ask for an explanation.2012-08-09
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    I like this, but it took me *quite a while* to grasp! (Of course, I'm rather low on the "totem pole"...)2012-08-09
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    @TheChaz I've added some color to help follow the logic.2012-08-09
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    Thanks, Bill. I didn't have a problem with this answer; it just took a little to change gears from the approach I had in mind (which **avatar** put into an answer) to this approach :)2012-08-09
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    @TheChaz Yes, that's probably the most straightforward way to proceed. But I thought it might prove of interest to emphasize that there is additional structure that simplifies matters. Such structure comes to the fore when one studies $q$-difference operators and related operator calculus.2012-08-09
7

Here $$f(3x+1)=3x(3x+1)=((3x+1)-1)(3x+1)$$ $$\implies f(x)=(x-1)x=x^2-x$$ $$\implies f(2x)-4f(x)=4x^2-2x-4x^2+4x=2x$$

In general, just let $3x+1=t$ and express $x$ in terms of $t$ and substitute in $f(t)$

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    Thanks, that's also what my theory says thanks for the solving the starting process :)2012-08-09
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    I did the math $$ x = ( w - 1 ) / 3 $$ and i have concluded to this $$ f(w) = w^2 - 2w + 1 + ( 3w - 3 ) / 3 $$ however there is an f(x) missing what should i do ??2012-08-09
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    $f(w)=w^2-2w+1+\frac{3(w-1)}{3}=w^2-2w+1+w-1=w^2-w$.Here, $w$ is just a dummy variable , so you can substitute $x$ for $w$ which gives $f(x)=x^2-x$. Keep in mind this new $x$ is different from previous $x$2012-08-09
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    +1 to both of you for the above comments (OP showing work and avatar following up)2012-08-09