Discrete Fourier Transform by hand
If you use the DFT formula, you get:
$$
Y[k] = \sum_{n=0}^{2N-1}y[n]e^{\frac{-2\pi k n}{2N}} 
$$
Now, substituting the definition of $y[n]$ you get:
$$
Y[k] = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k (2n)}{2N}} = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k n}{N}}
$$
So, for $0\leq k &lt; N$ you get that
$$ Y[k] = X[k]$$
and for $k\geq N$ you get
$$
Y[k] = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k n}{N}} = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi (k-N) n}{N}} = Y[k-N]
$$
Therfore, 
$$
Y[k] = \begin{cases}X[k] &amp; 0\leq k &lt; N \\ X[k-N] &amp; k\geq N\end{cases}
$$
or $X[k]=\{4,3,2,1,0,1,2,3,4,3,2,1,0,1,2,3\}$