0
$\begingroup$

I just saw this from a book. What is the real life application of it?

enter image description here

$\endgroup$
1
  • 3
    $\begingroup$ If the theorem is in a book, then keep reading the book and it should, at some point, be used... $\endgroup$ – 5xum Sep 21 '17 at 9:26
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.