I am confused by two exercises from a textbook.
According to the solution, both exercises can be solved by using the following equation for a discrete Fourier transform
$y_j = y(t_j) = a_0 + \displaystyle\sum_{k=1}^{n} a_k\cos\left(\frac{2\pi jk}{N}\right) + \displaystyle\sum_{k=1}^{n}b_k\sin\left(\frac{2\pi jk}{N}\right)$
with
$a_m = \dfrac{2}{N} \displaystyle\sum_{j=0}^{N-1}y_j\cos\left(\frac{2\pi jm}{N}\right)$
and
$b_m = \dfrac{2}{N} \displaystyle\sum_{j=0}^{N-1}y_j\sin\left(\frac{2\pi jm}{N}\right)$
See both exercises attached for more details.
Now my question: To derive $a_m$ and $b_m$ in exercise 18 they use $\sum_{j=1}^{N}$ instead of $\sum_{j=0}^{N-1}$. In exercise 19 they use the equation as indicated above. I can't see why there should be a difference between the two exercises. Is it because of the nature of the two data sets (one starting at $y=0$, the other at $y>0$)? Am I missing some theory/notation rules? Thanks for your help.
Edit: Is it due to the property of the first data set that data points are equally spaced? This would allow using
$a_m = \dfrac{2}{N} \displaystyle\sum_{j=1}^{N}y_j\cos\left(\frac{2\pi jm}{N}\right)$
and
$b_m = \dfrac{2}{N} \displaystyle\sum_{j=1}^{N}y_j\sin\left(\frac{2\pi jm}{N}\right)$