Confusion in Discrete Fourier Transform derivation

Question

In the book "Digital Image Processing" by R. Gonzalez, section 4.4.1 "Obtaining the OFT from the Continuous Transform of a Sampled Function",
they get $$\tilde{F}(\mu)=\sum^{\infty}_{n=-\infty}f_ne^{-j2\pi\mu n\Delta T}$$ where $f_n$ are the sampled points from a continuous function $f$ and $\tilde{F}$ is the fourier transform of the sampled function.
Now, the book says:
"Suppose that we want to obtain $M$ equally spaced samples of $\tilde{F}(\mu)$ taken over the period $\mu= 0$ to $\mu=\frac1{\Delta T}$. This is accomplished by taking the samples at the following frequencies: $$\mu=\frac{m}{M\Delta T} \text{ , } m=0,1,2,...,M-1$$ Substituting this result for $\mu$ into <first equation> and letting $F_m$ denote the result yields" $$F_m=\sum^{M-1}_{n=0}f_ne^{-j2\pi \frac{mn}M}\text{ , } m=0,1,2,...,M-1$$ Now what I don't understand is that when we are obtaining samples of $\tilde{F}$, then all we have to do is get values of the function at the given points. So how did doing just that, change the limits of the summation? According to me, $$F_m=\sum^{\infty}_{n=-\infty}f_ne^{-j2\pi \frac{mn}M}\text{ , } m=0,1,2,...,M-1$$ this should be the correct expression. So where did I go wrong? Please explain.

Answer 1

Perhaps you and the author are confusing the term given samples. I think the author means that $M$ values of $f_n$ are given. In this case the finite sum is correct. However you suppose that $M$ values of $F_m$ are given. Here the infinit sum is correct. Note that in both cases the result is not identical.