# What is the theorem of Bliss?

I just saw this from a book. What is the real life application of it? • If the theorem is in a book, then keep reading the book and it should, at some point, be used... – 5xum Sep 21 '17 at 9:26

First thing that comes to mind is this: assume there is continuous function $f$, of which you are asked to calculate numerically the Fourier coefficients over $[a,b]$.
However, you do not have the luxury of knowing the actual function $f$. The only things you know are a family $U=\{\sigma_1,\sigma_2,\cdots\}$ of subdivisions of the interval $[a,b]$ - say, $\sigma_j=\left(a,x_j^1,x_j^2,\cdots,x_j^{k_j-1},b\right)$ - which become finer and finer, and the values $$f_\sigma^s=\frac1{x_\sigma^s-x_\sigma^{s-1}}\int_{x_\sigma^{s-1}}^{x_\sigma^s} f(t)\,dt$$
Can you complete the task? Can you write an algorithm that does it for you? The answer is yes: since $f$ is continuous, by integral MVT there are $\xi_\sigma^s\in(x_\sigma^{s-1},x_\sigma^s)$ such that $f_\sigma^s=f\left(\xi_\sigma^s\right)$, and, by Bliss' theorem, $$\lim_{\sigma\in U}\sum_{k=0}^{k_\sigma} \left(x_\sigma^s-x^{s-1}_\sigma\right)g(y_\sigma^s)f_\sigma^s=\int_a^b g(t)f(t)\,dt$$ where the points $y_\sigma^s\in \left[x_\sigma^{s-1},x_\sigma^s\right]$ can be fixed in advance (depending only on $U$).
Of course, this specific instance could be done with other theorems, but the idea adapts to the more general case when $f_\sigma^s$ is just one value which $f$ takes in $(x_\sigma^{s-1},x_\sigma^s)$, rather than exactly the average. And not knowing exactly the value of a function at some point, but rather one at a very close point, is quite common in applications.