Confusion with a function transformation I got a HW problem wrong in my Signals and Systems class and am hoping someone can help me understand why. 
There's a discrete-time signal x[n] = u[n] + 2u[n-3] - 3u[n-6], where u[n] is the discrete-time unit step. The problem said to first draw this function, then draw the function transformation x[4n+1]. 
I drew the original function x[n] correctly, but got the transformation x[4n+1] wrong. I thought I should follow the order of operations, so first I compressed the signal by 4, and then I shifted it to the left by 1. Apparently I was supposed to shift it first, then compress it to get the correct answer. So my question is, why does this not follow the order of operations?

It might be easier to understand if we write the scaling and translation in two steps.
To that end, suppose the $f(n)$ is a given function.  Then for two numbers $a$ and $b$, we define $g(n)$ as the function given by $$g(n)=f(an)$$ and $h(n)$ as the function given by 
$$\begin{align}
h(n)&=g(n+b/a)\\\\
&=f\left(a(n+b/a)\right) \tag 1\\\\
&=f(an+b)
\end{align}$$
Now, let's see what happens upon evaluating these functions at various integer values of $n$.
For $n=0$, $h(0)=g(b/a)=f\left(a(0+b/a)\right)=f(b)$.  
For $n=1$, $h(1)=g(1+b/a)=f\left(a(1+b/a)\right)=f(a+b)$.  
For $n=2$, $h(2)=g(2+b/a)=f\left(a(2+b/a)\right)=f(2a+b)$.
We see now that $h$ is a transformed image of $f$ in which we first translate $f$ to the "left" by $b/a$ and then "compress" by a scale factor of $a$.  
This is different from first compressing by the factor $a$ and then translating by $b/a$.  In this case, we would obtain $f(an+b/a)\ne f(an+b)$.
