Sunday, 29 December 2019

definition - How do we define the sum of a bi-infinite series?




If we have a (usual) series of the form
S=n=1an
then we define S to be the limit as N goes to infinity of the partial sums SN=Nn=1an,
provided the limit exists. If we instead have a bi-infinite series of the form
S=n=an
then how do we define this sum? Is it
lim
or
\lim_{N \to \infty} \lim_{M \to \infty} \sum_{n = -M}^N a_n
or something else? Can you also refer me to any standard textbook that deals with bi-infinite series? Thanks!


Answer



Bi-infinite summation is summation over \Bbb Z. In general (I think that I read about this in Professor Tao's Analysis book), one defines a sum over an arbitrary set X as follows.



Let X be any set and F(X) be the set of all finite subsets of X. Let f: X \to \Bbb R be nonnegative.




\sum_{x \in X} f(x) := \sup\{\sum_{x \in A} f(x): A \in F(X)\}



Now if f is not necessarily nonnegative, then in case \sum_{x \in X} |f(x)| is finite,



\sum_{x \in X} f(x) := \sum_{x \in X} f^+(x) - \sum_{x \in X} f^-(x)



Where f^+(x)= \max\{f(x), 0\} and f^-(x) = \max\{-f(x), 0\}.



Otherwise \sum_{x \in X} f(x) is undefined.




In the case X = \Bbb Z, it's easy to check that when \sum_{x \in \Bbb Z} f(x) is defined, we have that the limit:



\lim_{N \to \infty} \sum_{n=-N}^N f(n)



exists and equals that thing.


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...