Extracting even / odd part of summation trick. Given a function f(x), we know its even part is given as (f(x)+f(-x))/2 and its odd part is given by (f(x)-f(-x))/2.

Maribel Vang 2022-10-28 Answered
Extracting even / odd part of summation trick
Given a function f(x), we know its even part is given as f ( x ) + f ( x ) 2 and its odd part is given by f ( x ) f ( x ) 2 .
Consider a discrete sequence given by a j for j 1, then the sum of the terms in the sequence till n terms is given as
S = j n a j
Suppose I wanted to get the sum of the even terms in the above expression; then
S o d d = j n a j + a j 2
and, for odd,
S o d d = j n a j a j 2
But wait, our sequence was defined for j 1. Well, here's the thing: a j is some function of j, extending the domain to negative integers and evaluating the function gives the right answer... but I can't understand why the continuous function trick extended to here.
Examples:
Sum of first n numbers given as n ( n + 1 ) 2 , sum of first n odds will be given as: n ( n + 1 ) 2 n ( 1 n ) 2
My attempt at finding an exact connection: To the discrete sequence n ( n + 1 ) 2 , we can associate a function f ( x ) = x ( x + 1 ) 2 and we can think of the summation as summing this function at several different input points i.e:
S = j n a j k n f ( x + k )
Then we apply the even odd decomposition and return back to the sequence world.
My question: Does there exist an association for every sequence with a function? If not, what is the criterion for an association to exist?
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Answers (2)

honejata1
Answered 2022-10-29 Author has 10 answers
Step 1
The definition of "even" and "odd" used in the function definition mean that for an "even" function g , g ( x ) = g ( x ) for any x, and for an "odd" function h , h ( x ) = h ( x ): the definitions are unrelated to the divisibility by 2 of any numbers.
Step 2
So how are the discrete sequences related? Well, sorry to burst your bubble, but they aren't. Your formula for odds working is a coincedence, and it doesn't work for evens (it evaluates to n, which is not the sum of the first n even numbers)
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Hugo Stokes
Answered 2022-10-30 Author has 7 answers
Step 1
A different approach yields the sum of the even terms:
S even = j a 2 j = j 1 + ( 1 ) j 2 a j
Step 2
Similarly, the sum of the odd terms is
S odd = j a 2 j 1 = j 1 ( 1 ) j 2 a j
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