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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: $$f(2x) - 4f(x) = 2x$$

M Turgeon
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Chris
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3 Answers3

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$$\begin{align} \overbrace{\dfrac{f(3x\!+\!1)}{3x\!+\!1} = 3x}^{\displaystyle \color{#c00}{\dfrac{f(z)}z\, =\, z\!-\!1}}\ \Rightarrow\ \dfrac{f(2x)\!-\!4f(x)}{2x} \,=\,&\ \color{#c64}{\dfrac{f(2x)}{2x}}-\, 2 \color{#C00}{\dfrac{f(x)}x} \\[.2em] =&\,\ \color{#c64}{2x\!-\!1} -\ 2(\color{#C00}{x\!-\!1})\\ =&\,\ 1\end{align}\qquad$$

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).

Bill Dubuque
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    Why would this be downvoted? – Pedro Aug 09 '12 at 19:12
  • I like this, but it took me quite a while to grasp! (Of course, I'm rather low on the "totem pole"...) – The Chaz 2.0 Aug 09 '12 at 19:16
  • @TheChaz I've added some color to help follow the logic. – Bill Dubuque Aug 09 '12 at 19:18
  • 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 :) – The Chaz 2.0 Aug 09 '12 at 19:20
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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. – Bill Dubuque Aug 09 '12 at 19:27
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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 $f(t)$

Aang
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  • Thanks, that's also what my theory says thanks for the solving the starting process :) – Chris Aug 09 '12 at 18:16
  • 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 ?? – Chris Aug 09 '12 at 18:40
  • $f(w)=w^2-2w+1+\frac{3(w-1)}{3}=w^2-2w+1+w-1=w^2-w$.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 $x$ – Aang Aug 09 '12 at 18:44
  • +1 to both of you for the above comments (OP showing work and avatar following up) – The Chaz 2.0 Aug 09 '12 at 19:17
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Hint: $3x+1=y \Rightarrow x= \frac{y-1}{3}$. So replace $x$ by $\frac{y-1}{3}$ and magic happens.

N. S.
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