periodic function, find g(6) 
If $f$ is periodic, $g$ is polynomial function, $f(g(x))$ is periodic, $g(2)=3$, and $g(4)=7$, then $g(6)$ is
A) 13
B) 15
C) 11
D) none of these

The answer is c) 11, but I did not understand how $g(x)$ was considered linear polynomial (because i got answer when $g(x)$ is $2n -1$), isn't there any $g(x)$ with degree greater than 1 make $f(g(x))$ periodic?
Why? How will you solve the problem?
 A: Let $g(x)$ be a polynomial, that is: $$g(x)=\sum_{n=0}^\infty c_nx^n$$
with every $c_t$ a constant, note a polynomial of degree $k$ has $k=\max\{t|c_t\neq 0\}$
We take that:
$$\forall x\in\Bbb R,\forall t\in\Bbb Z; f(x)=f(x+t\alpha)$$
$$\forall x\in \Bbb R, \forall t\in\Bbb Z; f(g(x))=f(g(x+t\alpha))$$
Note $t$ is a constant independent of $x$.
We then insert our sum form of the polynomial.
$$f(\sum_{n=0}^\infty c_nx^n)=f(\sum_{n=0}^\infty c_n(x+\alpha)^n)$$
Expanding gives:
$$f(\color{red}{c_0}+\color{purple}{c_1x}+\color{blue}{c_2x^2}+\color{green}{c_3x^3}+...)=f(\color{red}{c_0}+\color{purple}{(c_1x+c_1\alpha)}+\color{blue}{(c_2x^2+2c_2x\alpha+c_2\alpha^2)}+\color{green}{(c_3x^3+3c_3x^2\alpha+3c_3x\alpha^2+c_3\alpha^3)}+...)$$
Notice when $k$ (as defined earlier as the degree of the polynomial) is at least $2$ the $t$ which satisfies $f(g(x))=f(g(x+t\alpha))$ is dependent on $x$. This is the same thing as saying that $(x+\alpha)^n-x^n-\alpha^n$ has terms for $n\geq 2$, which is trivial by the binomial theorem.
$$(x+\alpha)^n-x^n-\alpha^n=\sum_{r=1}^{n-1}\bigg[\binom nr x^{n-r}\alpha^r\bigg]$$

NB when $k=1$, you get:
$$f(c_0+c_1x)=f(c_0+c_1x+c_1\alpha)$$
so $t=c_1$, a non-zero constant independent of $x$
A: If $g$ is a polynomial one shouldn't assume $g(x)$ is periodic and so the condition that $f(g(x))$ is periodic is quite unusual.
If $g(x)=mx+b$ is linear, and $p$ is the period of $f$ it's easy to see that $f(g(x+\frac pm)) = f(mx +p + b)=f(g(x) + p) = f(g(x))$ and $f(g(x))$ is periodic.
If $g$ isn't linear can we still have $f(g)$ be periodic?  Not really.
This is a bit handwavey but I hope it is intuitively clear.
Let's suppose the period of $f$ is $m$.  So $f(x+m) = f(x)$.  Now if $k$ is not a muliple of $m$ that does not mean that $f(x+k) \ne f(x)$ but in general that will not be the case.  And if you view all potential values $x$ you'd find $f(x+k) \ne f(x)$ "almost everywhere".
Now if $f(g(x))$ has a period of $h$ then $f(g(x+h)) = f(g(x))$ and if $g(x+h) - g(x) = k$ then $f(g(x) + k) = f(g(x))$ which means that $g(x+h) - g(x)$ is a multiple of $m$ "almost everywhere".
Okay but if there is a $g(x)$ where $g(x+h) = g(x) + km$ for some multiple $km$ or $m$ and $g$ is not linear then we can find some some $e > 0$ where $g(x+e)- g(x) \ne g((x+h) + e)-g(x+e)$ (because $g$ is not linear) and were $g(x+e)-g(x) < m$ and $g(x+h+e) -g(x+h) < m$ (because $e$ can be arbitrarily smalland polynomials are continuous).
And that means that the difference $g((x+e) +h)$ and $g(x+e)$ is not a multiple of $m$ and that in general "almost nowhere" does $f(g(x+e+h)) = f(g(x))$.
Okay, that was really hand wavey and Rhys Hughes proof is better but I think for the problem my type of thinking (or in similar terms) isn't meant to be proven but taken to be intuitively "obvious".
Frankly I don't like this problem....
A: $g(x) = 2x-1$ is still a polynomial, it's just a little atypical in that it's a very simple polynomial, which has no degrees greater than 1. That still fits the definition, it's just not what you'd typically think of when you hear the term "polynomial".
