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# Cyclic functions.

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I have the problem :

Find all functions $$f:\mathbb R \to \mathbb R$$ that satisfy

$$f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x$$

for all nonzero $$x$$. Note that the domain of $$f$$ is all real numbers. Make sure you define your function everywhere!

Hint: If you substitute $$(x - 1)/x$$ in for $$x$$, then you get a new equation that also involves $$f\left( \frac{x - 1}{x} \right)$$Then do this substitution again and you may get another useful equation to consider.

I think this is related to the fact that $$f(x) = \frac{x - 1}{x}$$ is cyclic with order 3, but I don't know how to use this to find the answer.

Nov 21, 2018

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That proof is incomplete. The problem needs f(x) to be defined for all x, but in the proof the function is not defined for x=0,1.

Guest Nov 21, 2018
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Ok, Guest, I'll see if I can help you.

PartialMathematician  Nov 21, 2018
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StackExchange...huh...I answered some questions there about computer science. I didn't know they also had a math webpage.

Nov 21, 2018
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In the proof, 0 is restricted because the denominator of a fraction cannot be 0 (cannot divide by 0). If we set x to 1, then the part $$3f(\dfrac{x-1}{x})$$ cancels out and turns into 0 because $$3f(0) = 0$$. Also, if you look around in the proof, you will find $$x$$  and $$1-x$$ many times as the denominator, which cannot equal 0. Simply stated, 0 and 1 are restricted, AKA not in the domain of the function.

Hope this helps,

- PartialMathematician

Nov 21, 2018
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Here's my take on this:

Nov 21, 2018